Wolfram Alpha wrong answers on $(-8)^{1

Wolfram Alpha wrong answers on $(-8)^{1 Техника

In the event that $\rho$ is an integer, we can create a definition for $c^\rho$ for all real values of $c$, and then proceed with solving for $c$. If $\rho$ is even, $c^\rho$ will always be greater than or equal to zero, prohibiting the solution of $c^\rho=x$ for negative values of $x$. However, if $\rho$ is odd, negative values of $x$ are solvable.

When we’re using that, it is
true by definition
that $\sqrt[3]{a^3}=a$ for all real $a$.
However, Wolfram Alpha is not smart enough to notice that you probably want the real cube root, so it uses a different function that gives complex outputs for negative real inputs, but on the other hand is continuous on in the complex plane
from the real axis.
First let $x^3=1$, thus
$x^3-1=0$, now convert this to the form $(x-1)(x^2+x+1)=0$ solving first bracket will give you real cube root and the second one will give the imaginary cube roots.

Table of contents

  • Wolfram Alpha wrong answers on $(-8)^{1/3}$ and more? [duplicate]
  • Is the cube root of $a^3$ always a
  • How can I calculate all results (3) of a cube root?
  • Evaluating Cube roots of fractions
  • What is the cube root of-8 in Wolfram’s equation?
  • What is the cube root of X?
  • Is it possible to get negative cube roots in WolframAlpha?
  • How to find the root of a cubic root?

Wolfram Alpha wrong answers on $(-8)^{1/3}$ and more? [duplicate]


Wolfram Alpha doesn’t give $-2$ for $(-8)^{1/3}$, and it absolutely fails to draw $f(x)=x^{1/3}$ — does anyone know why? Am I missing something very ‘deep’ Wolfram Alpha is trying to teach me?

Here’s what the graph should look like:
Wolfram Alpha wrong answers on $(-8)^{1

And here’s what Wolfram Alpha draws:
Wolfram Alpha wrong answers on $(-8)^{1

Solution 1:

What you are seeing is that every non-zero number has three distinct «cuberoots». In the first picture you’ve posted, the software very nicely graphed the real cuberoots. The problem is that Wolfram|Alpha is drawing a different branch of the cuberoot than you are used to, hence the labels of real and imaginary parts on the graph.

Solution 2:

The principal value of the cube root is given by



Wolfram Alpha

. If you really just want the real value, input with

Sign[x] Abs[x]^(1/3)

Solution 3:

What you are probably thinking about is that for solving the equation $$z^k = a$$

we often have many solutions (for $z$) in the complex plane. More to the point, for a


one wants to

assign one out-value

for each


. Choosing a particular way to assign function values is called

picking a branch

for a function. For some functions there exist widely used conventions how to pick these branches and they get called


branch. Integer roots and logarithms are famous examples of functions which have a

principal branch


Evaluating Cube roots of fractions, $\begingroup$ Regarding the typesetting of the formulas, Wolfram Alpha is very good at guessing what you mean when you type a formula, but it doesn’t give …

Cube root by long division

We all know how to find square


by long division. But do you have heard of

cube root

by long division. If no, then you must see this video till the end.

Is the cube root of $a^3$ always a


My question is as follows: for any real number a, is the cube root of the cube of a always equal to a?

For a = 2, the result seems to hold. However for a = -1, wolfram alpha says the cube root of -1 is non real? But,

$-1 = -1\times-1\times-1$

right? So surely the cube root of -1 is -1 and then shouldn’t this also fit the statement: is the cube root of the cube of a always equal to a?

Perhaps I am confusing myself somewhere and there is something silly I am missing.

Thanks 🙂

Solution 1:

There are different possible «cube root» functions. They give the same result when the input is on the positive real axis, but differ elsewhere.

As long as we’re considering only real inputs, the natural cube root function is simply the inverse function of $x\in\mathbb R \mapsto x^3$. When we’re using that, it is

true by definition

that $\sqrt[3]{a^3}=a$ for all real $a$.

However, Wolfram Alpha is not smart enough to notice that you probably want the real cube root, so it uses a different function that gives complex outputs for negative real inputs, but on the other hand is continuous on in the complex plane


from the real axis.

You will notice that the result it gives you indeed is


cube root of $-1$: $\left(\frac 12 + \frac{\sqrt 3}{2}i\right)^3 = -1$

Solution 2:

The cube root of -1 is certainly -1.

For real numbers, you can only take the square root of a number greater than or equal to 0, but you can take the Cube Root of Any Number.

Solution 3:

Over the real numbers, each number has exactly one cube root and the cubed root of a number cubed is itself. But over the complex numbers, each real number has exactly 3 cube roots and only one of those is real. Wolfram Alpha is probably considering the general case of complex numbers.

Solution 4:

The fundamental theorem of algebra states that a polynomial of degree $n$ has at least $n$ roots up to multiplicity over an algebraically closed field.

In simpler terms, it means that the polynomial $a^3 = -1$ has at least three roots.

One of these roots is $-1$: $(-1)^3 = -1$.

However, it also has two other roots that are complex; specifically, the rotation of $-1$ about the origin in the complex plane by $\frac{2\pi}{3}$ radians, twice. So that’s how you get the other two roots.

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For if $z = \left[\exp i\frac{5\pi}{3}\right]^3$, then $z^3 = \exp i\frac{3\cdot 5\pi}{3} = \exp 5i\pi = \exp i\pi = -1$, and so on.

Algebra precalculus — cube root simplification, As a result, the norm of a cube is the cube of the norm, and the norm of a cube root is the cube root of the norm. In your case, the norm of …

How can I calculate all results (3) of a cube root?


According to Wikipedia and Wolfram Alpha, a cube root $n^{\frac{1}{3}}$ has three results: one real number and two complex, if $n$ is a real number; and three complex numbers if $n$ is a complex.

Most of calculators returns only the principal root, but not the other two results. Is there a way to calculate the two remaining?

Solution 1:

Just multiply your root by $e^{2i\pi/3}$ and $e^{-2i\pi/3}$.

For example : solve $x^3=8$. A trivial root is 2, therefore the roots will be
2\, e^{2i\pi/3}\quad
2\, e^{-2i\pi/3}.
If you want to solve $x^3=-i$, a simple root is $e^{-i\pi/6}$ (this because $-i=e^{-i\pi/2}$), therefore the roots are
e^{-i\pi/6}\, e^{2i\pi/3}= e^{i\pi/2}\quad
e^{-i\pi/6},\ e^{-2i\pi/3}= e^{-5i\pi/6}.

Solution 2:

Yes, there is a way .e.g. First let $x^3=1$, thus
$x^3-1=0$, now convert this to the form $(x-1)(x^2+x+1)=0$ solving first bracket will give you real cube root and the second one will give the imaginary cube roots.

Find complex roots of $x^2+x+1$, If x is a real number, then, unless one carefully explains otherwise (which WolframAlpha did not), x 3 is defined as the unique real cube root of x, so …

Evaluating Cube roots of fractions


For example : $\sqrt[3]{\frac {21}{37}}$
If I evaluate it in a scientific calculator, the answer would be $0.8280 \dots$. But is there another way to solve it instead of using the calculator? And why the answer just had to be in decimals? Can’t the answer be in fractions?

Solution 1:

Well, most calculators output decimals instead of symbolic expressions, so it’s not surprising that that’s what you got. Symbolically, there’s not much that you can do… you can write it as $\frac{\sqrt[3]{21}}{\sqrt[3]{37}}$ if you wish, but I’m not convinced that’s a meaningfully better way to write it. This article details one method for computing cube roots by hand.

Solution 2:

First, you won’t be able to write the answer as a fraction because if $(a/b)^3 = 21/37$, then we have $37a^3 = 21b^3$. Therefore the numbers $37a^3$ and $21b^3$ must have the same prime factorization. But the number of times $7$ appears in the prime factorization of $37a^3$ is a multiple of $3$, whereas the number of times it appears in that of $21b^3$ is one more than a multiple of three.

Now to calculate a few decimal places, we use the fact that $21/37 \approx (5/6)^3$. Specifically,
$$\sqrt[3]{\frac{21}{37}}= \frac{5}{6}\sqrt[3]{1-\frac{89}{4625}}.$$

For $n \geq 1$ and $0 < x < 1$, taylor’s formula gives us
$$(1-x)^{1/3} = 1 — \frac{1}{3}x + \binom{1/3}{2}x^2 — \dots +(-1)^{n-1}\binom{1/3}{n-1}x^{n-1} + \varepsilon_n(x),$$
where the (necessarily negative) error term $\varepsilon_n(x)$ satisfies the inequalities
$$(-1)^n \binom{1/3}{n}x^n (1-x)^{1/3 — n} \leq \varepsilon_n(x) \leq (-1)^n \binom{1/3}{n}x^n.$$
(See Theorem 7.7 in

Calculus, Vol. 1,

by Apostol.) We can weaken this inequality slightly to the more convenient form
$$(-1)^n \binom{1/3}{n}\left(\frac{x}{1-x}\right)^n \leq \varepsilon_n(x) \leq (-1)^n \binom{1/3}{n}x^n.$$

Now say we want to know the value of our cube root to within $10^{-9}$. We look for a value of $n$ such that the left-hand side in the above inequality is less than $10^{-9}$ for $x = 89/4625$. We have $x/(1-x) < 0.02$, so $(-1)^5\binom{1/3}{5}[x/(1-x)]^5 < (22/729)(2 \times 10^{-2})^5$, which is less than $(6/5) \times 10^{-9}$, showing that $n = 5$ will provide enough precision.

Returning to Taylor’s formula, we now have
$$\frac{5}{6}\left[1 — \frac{1}{3}x — \frac{1}{9}x^2 — \frac{5}{81}x^3 — \frac{10}{243}x^4 — \frac{22}{729}\left(\frac{x}{1-x}\right)^5\right] \leq \sqrt[3]{\frac{21}{37}}\leq \frac{5}{6}\left[1 — \frac{1}{3}x — \frac{1}{9}x^2 — \frac{5}{81}x^3 — \frac{10}{243}x^4 — \frac{22}{729}x^5\right], $$
where $x = 89/4625$, yielding
$$ 0.827953329397 \leq \sqrt[3]{\frac{21}{37}} \leq 0.827953329405 $$
or an accuracy better than $10^{-11}$. Compare this with the actual value $\sqrt[3]{21/37}= 0.827953329403…$

Granted, this method is not easy to carry out without a calculator, but it shows how such a calculation can be done.

Solution 3:

In the same spirit as user49640’s answer, knowing that the solution is close to $\frac 56$, we could build the simplest Pade approximant of function $x^3-\frac{21}{37}$ around $x=\frac 56$.

This would give $$x^3-\frac{21}{37}\approx \frac{\frac{89}{7992}+\frac{6893 }{3330}\left(x-\frac{5}{6}\right)}{1-\frac{6}{5}
\left(x-\frac{5}{6}\right)}$$ from which the solution is $$x=\frac{68485}{82716}\approx 0.827953$$

Another way would be using Newton method for which the iterates would be
n & x_n & \text{approx} \\
0 & \frac{5}{6} & 0.833333 \\
1 & \frac{6893}{8325} & 0.827988 \\
2 & \frac{982489486039}{1186648388775} & 0.827953

Solution 4:

If the numbers in the cube root are themselves perfect cubes, eg: $\sqrt[3]{\frac{343}{27}}=\frac{7}{3}$, as $7^3=343$ and $3^3=27$ or you can rewrite them as perfect cubes, then you’ll be able to get exact answers as fractions or integers out of them. However, more often than not most numbers aren’t well behaved like this, and the best you can do is an approximation.

For calculators it’s a lot easier to compute decimals instead of fractions so that’s what they do, but if you wanted to do a fractional approximation, you could use a power series of $\sqrt[3]{x}$ centered at $c$:

Where you’d select $c$ as the closest Perfect Cube larger than $x$, where $x$ is the number you want to approximate. To actually compute it by hand instead of letting the sum range from $0$ to $\infty$, just cap it off and calculate the first couple terms. The more terms you calculate the better the approximation obviously. For example to find $\sqrt[3]{24}$, you’d let $x=24$, and $c=27$, since $27$ is the next largest perfect cube bigger than 24:
And then just start calculating terms in the sequence, starting by plugging in $n=0$, then $n=1$, $n=2$, and so on. In this case you would get:
It’s important to note that this is just an approximation, and that the root isn’t really equal to the fraction. But since every other answer here, and really even what the calculator spits out are just approximations as well, I guess it doesn’t matter. The main difference is that this approximation will give you fractions, as opposed to the standard decimal methods.

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Why is it wrong to take the cube root of both sides, 3 Answers Sorted by: 6 You can take cube roots, provided you take into account the fact that you’re allowing for complex solutions. x 3 = 1 has three …

To recover the third solution in Mathematica as is to explain why it is missing in the solution without extra conditions.

zeros=Reduce[(3 - Cos[4 x]) (Sin[x] - Cos[x]) == 2 && x >= -2 π && 
   x <= 0, x] /. C[1] -> 0

So the solution implicitly is the substitution for C1==0. This has to done by interpreting the periodicity of the trigonometric composition.

FunctionPeriod[(3 - Cos[4 x]) (Sin[x] - Cos[x]), x]

That is too what the constant C1 is for in the solution of Solve and Reduce.

So getting the -Pi in the solution was a different interpretation in the question and is not an error.

This shows each period of the functions has two zero on the real axis.

Solve[(3 - Cos[4*x])*(Sin[x] - Cos[x]) == 2, x]

output of Solve for the first question

This shows the constant C1 and this domain of C1. So there are as many solutions as the Integers are mighty.

The ConditionalExpression is much to advanced for beginners. In the Plot built-in it does not matter because Plot uses Through. That evaluates the ConditionalExpression on the choosen interval of the Plot. In the output of Mathematica, the substitution evaluates them too and this makes a choice of an interval in which the solution reside. Sinces most beginners courses in maths do not teach Complexes there are obsolete solutions.

zlist = List @@ zeros[[All, 2]]
zlist = Append[zlist, zlist + 2 π] // Flatten

Plot[(3 - Cos[4 x]) (Sin[x] - Cos[x]) - 2, {x, -2 Pi, 2 Pi}, 
 Ticks -> {{-2 Pi, -Pi, 0, Pi, 2 Pi}, {-1, 1}}, 
 Epilog -> {Red, PointSize[0.01], Point[Thread[{#, 0} &@zlist]]}]

Plot of the function and the zeros

 ComplexPlot[(3 - Cos[4*x])*(Sin[x] - Cos[x]) - 
  2, {x, -2 π - 2 π I, 2 π + 2 π I}, 
 Epilog -> {Red, PointSize[0.01], Point[Thread[{#, 0} &@zlist]]}]

ComplexPlot and the zeros

For the second part of the question

Reduce[Tan[2 x] Tan[7 x] == 1, x, Reals]


FunctionPeriod[Tan[2 x] Tan[7 x], x]

Reduce is slightly easier to understand in the output compared to Solve.

Reduce contains Solve complete and the methods of Solve can restricted in Reduce with the option Method->Reduce Solve behave almost like Reduce. Instead of Solve Reduce is invoked.

In the documentation page of Reduce Wolfram Inc states that «For transcendental equations, Solve may not give all solutions:». On the other side «Reduce does not solve equations that depend on branch cuts of Wolfram Language functions:». So plot the function first and identify the problem.

Solve has much more options than Reduce in the documentation page. It is up to oneself to test their functionality in Reduce.

solt = Solve[Tan[2 x] Tan[7 x] == 1, x, Reals]
zert = solt[[All, 1, 2]] /. C[1] -> 0 // List // Flatten
plo = Plot[Tan[2 x] Tan[7 x] - 1, {x, -1.0125 π, 1.0125 π}, 
  Epilog -> {Red, PointSize[0.02], Point[Thread[{#, 0} &@zert]]}]

Plot of the given functions and the zeros

lip = ListPlot[
  Callout[{# // N, 0}, #, 
     LeaderSize -> {{32, 135 \[Degree], 6}, {5, 180 \[Degree]}}] & /@ 
   zert, PlotMarkers -> Automatic, PlotTheme -> "Web"];

Show[plo, lip, ImageSize -> 600, AspectRatio -> 1/4]

Annotated zeros for Tan[ 2x] Tan[ 7x] - 1

The substitution of the trigonometrics is a nice alternative. Both factors work brilliant and give a polynomial of fifth order.

equ1 = (3 - Cos[4 x]) (Sin[x] - Cos[x]) - 2 == 0;
equ2 = t == 3 - Cos[4 x];
Eliminate[TrigExpand[{equ1, equ2}], x]

(* 16 t^2 — 4 t^4 + t^5 == 32 *)

Solve[16 t^2 - 4 t^4 + t^5 == 32, Reals]
Plot[{16 t^2 - 4 t^4 + t^5, 32}, {t, -3, 3}, 
 Epilog -> {Red, PointSize[0.02], Point[{2, 32}]}]


Reduce[-Cos[x] + Sin[x] == 1, x, Reals]
Solve[-Cos[x] + Sin[x] == 1, x, Reals]

(Element[C[1], Integers] && x == Pi/2 + 2*Pi*C[1]) || 
  (Element[C[1], Integers] && x == Pi + 2*Pi*C[1])

{{x -> ConditionalExpression[Pi/2 - 4*Pi*C[1], Element[C[1], Integers]]}, 
  {x -> ConditionalExpression[2*(-(Pi/2) + 2*Pi*C[1]), Element[C[1], Integers]]}, 
  {x -> ConditionalExpression[2*(Pi/2 + 2*Pi*C[1]), Element[C[1], Integers]]}, 
  {x -> ConditionalExpression[Pi/2 - 2*(Pi + 2*Pi*C[1]), Element[C[1], Integers]]}}

Reduce[3 - Cos[4 x] == 2, x, Reals]

Element[C[1], Integers] && x == (Pi*C[1])/2

Table[(Pi*C)/2, {C, -6, 6}]

(* {-3*Pi, -((5*Pi)/2), -2*Pi, -((3*Pi)/2), -Pi, -(Pi/2), 0, Pi/2, Pi, (3*Pi)/2, 2*Pi, 
  (5*Pi)/2, 3*Pi} *)

The cause is 4 is even and this means the zeros of both factors match.

Excuse my lack of knowledge and expertise in math,but to me it would came naturally that the cubic root of $-8$ would be $-2$ since $(-2)^3 = -8$.

I came to trust Wolfram Alpha so I thought I’d ask you guys, to explain the sense of that to me.

Mark McClure's user avatar

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asked Mar 7, 2011 at 14:24

fasseg's user avatar

answered Mar 7, 2011 at 14:37

Mike Scott's user avatar

Mike Scott

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Although it’s been two years since this question was asked, some folks might be interested to know that this behavior has been modified in WolframAlpha. If you ask for the cube root of a negative number, it returns the real valued, negative cube root. Here, I just asked for «cbrt -8», for example:

enter image description here

Note «the principal root» button. That allows you to toggle back to the original behavior. Near the bottom, we still see information on all the complex roots.

enter image description here

We can plot functions involving the cube root and solve equations involving the cube root and it consistently acts real valued. If you just type in an equation, it will solve it, plot both sides and highlight the intersections. Here’s «cbrt(x)=sin(2x)»

enter image description here

answered Apr 26, 2013 at 20:00

Mark McClure's user avatar

Mark McClure

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See this. In particular, the prinicipal cube root has nonzero imaginary part.

answered Mar 7, 2011 at 14:36

NebulousReveal's user avatar

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Of course, you’re absolutely right about $-2$ being a cubic root of $-8$.

The point might be that there are actually, three different cubic roots of $-8$, namely the roots of the polynomial $x^3+8$. One of this roots is real ($-2$), the other two are complex and conjugate of each other.

answered Mar 7, 2011 at 14:42

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Andrea Mori

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Community's user avatar

answered Dec 2, 2016 at 2:43

dxiv's user avatar

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Wolfram is using the polar complex form of -8 = 8cis(π)
Then the cube root of this is 2cis(π/3), which is 1 + i√3 (an alternate form on Wolfram)

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Incidentally, if you take $(1 + i\sqrt3)^3$, you will get -8!

answered Mar 7, 2011 at 14:36

The Chaz 2.0's user avatar

The Chaz 2.0The Chaz 2.0

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answered Jan 9, 2015 at 15:32

JPH's user avatar

I recently came across an interesting discrepancy involving the cube root function.

Cube root

In the search box, I put «cube root of x», and it stated the «Result» was correctly written as {\sqrt[{3}]{x}}

This graph is the reflection of the graph y = x3 in the line . They are inverse functions.

We know that that cube root of a negative number is negative, so for example {\sqrt[{3}]{-{6}}}=-{1.817}, and we can see this makes sense on the graph above.

Assuming «cube root of» is the real-valued root.

There is an option to see the «principal root», but this just gave the same result.

Raising to the power 1/3

We learn early in the study of roots and fractional powers that we can write roots in terms of fractional exponents. In general, that means:


So for square root we have:


and for cube root:

So we would expect the graph for {y}={x}^{{1}\//{3}} to be the same as the one for {y}={\sqrt[{3}]{x}}.

The blue curve is labeled «real part» and the red curve is «imaginary part».

Curiously, the «Input» value is stated as: {\sqrt[{3}]{x}}, but that’s not actually what I entered. So part of the answer regards {\sqrt[{3}]{x}}={x}^{{1}\//{3}}, but the rest of the answer does not.

We know from the section on complex roots (see especially Exercise 4 at the end), that a cubic equation will have 3 roots (just as a quadratic equation has 2 roots). These 3 roots may be all real, or a mix of real and complex roots.

24x7 Tutor Chat

Example: What are all the cube roots of −8?

I zoomed out a bit to get this graph, and added some guide segments (in green):

Using the same thinking as Exercise 4 mentioned earlier, the complex solutions for x3 = −8 should be apart, giving (where {j}=\sqrt{-{1}}):

x = −2

x = 1 + 1.73j

x = 1 − 1.73j

The above graph does give us one of these solutions (the middle one, since we can see the real part is and the imaginary part is ), but it doesn’t give the other two solutions.

Once again, the page tells us it is assuming the «principal root», and gives us the option to choose the «real-valued root». If we do that this time, we get the real root only version, looking like the graph at the top of the page.

Scientific Notebook answer

Bloodhound car

The blue graph is {y}={\sqrt[{3}]{x}}, and Scientific Notebook gives the full real solution (in first and third quadrants), while the magenta (pink) graph, {y}={x}^{{1}\//{3}} is in the positive quadrant only.

Geogebra and Desmos answers

Both Geogebra and Desmos give the same «full real value» graph for both {y}={\sqrt[{3}]{x}} and {y}={x}^{{1}\//{3}}.

cube root

Similar issue to square root

I’ve written before about the number of solutions for √16. The answer there is one solution of course, whereas if you are asked to solve {x}^{2}={16}, you will get 2 solutions.


Don’t just accept the computer’s word for it when it gives you a graph, or the solution for some equation. It’s doing its best to figure out what you want to know, but can’t be expected to know the full context of your query, or necessarily give you all the possible answers.

24x7 Tutor Chat

See the 6 Comments below.

Get the full list of complex nth roots of a number


Basic Examples  

Find the three third roots of unity:

Find the sixth roots of 2:

Find the fifth roots of :

Properties and Relations  

Exact values are returned for exact input:

Use the resource function ComplexRootQ to confirm the output of :

Version History

  • – 24 January 2020

  • – 06 September 2019

  • – 23 August 2019

  • – 10 July 2019

License Information

This work is licensed under a

Creative Commons Attribution 4.0 International License

One way to access these new functions is to select the “Use the real-valued root instead” option below the input bar as shown in the previous example. Alternatively, we can access Mathematica’s CubeRoot function, for example, by typing “cube root” instead of using the power ^(1/3), as in the previous example.

Is a cube root of 8?

As the cube root of 8 is a whole number, 8 is a perfect cube….Cube root of 8 in radical form: ∛8.

1.What is the Cube Root of 8?
3.Is the Cube Root of 8 Irrational?
4.FAQs on Cube Root of 8

What is the cubed root formula?

The formula of cube root is a = ∛b, where a is the cube root of b. For example, the cube root of 125 is 5 because 5 × 5 × 5 = 125.

What is the cube root of 216 by prime factorization method?

Therefore, the cube root of 216 by prime factorization is (2 × 2 × 2 × 3 × 3 × 3)1/3 = 6.

What is the cube root for 216?

As the cube root of 216 is a whole number, 216 is a perfect cube….Cube root of 216 in radical form: ∛216.

1.What is the Cube Root of 216?
3.Is the Cube Root of 216 Irrational?
4.FAQs on Cube Root of 216

How do you find the cube root of 216?

The cube root of 216 is the number which when multiplied by itself three times gives the product as 216. Since 216 can be expressed as 2 × 2 × 2 × 3 × 3 × 3. Therefore, the cube root of 216 = ∛(2 × 2 × 2 × 3 × 3 × 3) = 6.

What is the cube root of 3 27?

The value of the cube root of 27 is 3. It is the real solution of the equation x3 = 27….Cube root of 27 in radical form: ∛27.

1.What is the Cube Root of 27?
3.Is the Cube Root of 27 Irrational?
4.FAQs on Cube Root of 27

What is the principal root of 3 216?

Hence, here 216 is supposed to be the volume of the cube and we need to find the value of its side, by taking its cubic root. Let, ‘n’ be the value obtained from 3√216, then as per the definition of cubes, n × n × n = n3 = 216….How to Find Cube Root of 216.

Number (n)Cubes (n3)

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