# Simplify the cube root [closed]

How would one simplify the cube root of $x^{64}$? Please show the solution step by step if possible. Been a while since I took math so please forgive the simplicity of this question... Thank you in advance.

• One reason for this question would be to emphasise that the answer is not $x^4$ (which some might be trapped into thinking by $4^3=64$) Commented Jan 21, 2018 at 8:58

$\sqrt[3]{x^{64}}=\sqrt[3]{x^{21\cdot 3+1}}=x^{21}\sqrt[3]{x}$.

• You have not taken into account what happens when $x$ is negative. Commented Jan 21, 2018 at 4:56
• Is that a valid reason for a downvote? Commented Jan 21, 2018 at 4:56
• @JMoravitz The question is about a cubic root, not a square root, and $\sqrt[3]{x^3}=x$ independently on the sign of $x$. Commented Jan 21, 2018 at 5:05
• @JMoravitz This is marked precalculus. Clearly we are not working with complex numbers. Commented Jan 21, 2018 at 5:09
• @AlexRavsky the principal $n$'th root of a number is the $n$'th root whose argument is closest to zero. Writing a number in polar form $re^{i\theta}$ where $r$ is non-negative real and $0\leq\theta<2\pi$, one has $\sqrt[n]{re^{i\theta}}=\sqrt[n]{r}\cdot e^{i\theta/n}$ where $\sqrt[n]{r}$ is defined as usual for non-negative real numbers. See $n$'th root on wikipedia and $n$'th complex roots Commented Jan 21, 2018 at 5:43

$x^{64}=x^{\frac{64}{3}\times3}$.

Hence, $^3\sqrt{x^{64}}=x^{\frac{64}{3}}$.

Even when $x$ is negative, it yields the same results at there is an even power of $64$ which cancels out the negative.

On the other hand, the similar looking $x^\frac{5}{3}$ does not hold for negative values as $5$ is a odd power, and hence the negative sign is not cancelled.

Hence, we can form a generalisation for all negative $x$: for $x^\frac{a}{3}$, if $a$ is even, the result is positive. If $a$ is odd, the result is negative.

• You have not taken into account what happens when $x$ is negative. Commented Jan 21, 2018 at 4:56
• it is the same... the power 64 cancels the negative Commented Jan 21, 2018 at 5:01
• mmm... no. If you want to be pedantic, when you use non-integer rational exponents you actually get a multivalued expression, or if you want to make it single-valued you use the denominator of the rational number first rather than the numerator, specifically using the principal root. Your expression $x^{\frac{64}{3}}$ under such a convention yields a nonreal complex number for $x=-1$ Commented Jan 21, 2018 at 5:14
• i tried $x=-1$ and got 1. Commented Jan 21, 2018 at 5:16
• @XcoderX You should really elaborate some more on why $^3\sqrt{x^{64}}=x^{\frac{64}{3}}$ may be valid here, while the very similarly looking $\sqrt[6]{x^{10}}=x^{\frac{5}{3}}$ is clearly not.
– dxiv
Commented Jan 21, 2018 at 5:22

You need to be careful. Assuming that $x$ is real, one has $x^{64}$ is always a non-negative real number, and the cube root of a non-negative real number will again be a non-negative real.

$\sqrt[3]{x^{64}}=|x|^{\frac{64}{3}}=|x^{21}\cdot \sqrt[3]{x}|$

Without such care, one can arrive at incorrect results such as $\sqrt{x^2}=x$, which would imply things such as $1=-1$ (by plugging in $-1$ in for $x$)

• As Alex Ravsky has pointed out, for odd powers this is unnecessary. The equation $\sqrt[3]{a^3} = a$ holds for all real $a$. Commented Jan 21, 2018 at 5:07
• I am not taking issue with $\sqrt[3]{a^3}=a$. I am taking issue with all of the other erroneous simplifications being made @YakovShklarov. Commented Jan 21, 2018 at 5:10
• @E.H.E. $\sqrt{x^2}=|x|$ and not $\sqrt{x^2}=x$ is a classic example of why care needs to be taken in such things which one can see in problems such as Why $\sqrt{-1\times -1}\neq \sqrt{-1}^2?$. And I take back what I said earlier. I do take issue with saying $\sqrt[3]{a^3}=a$. That would be true if we were using the realvalued cube root rather than the more standard convention of using the principal cube root where such a simplification is false. $\sqrt[3]{(-1)^3}=\sqrt[3]{-1}=\frac{1}{2}+\frac{\sqrt{3}}{2}i$ Commented Jan 21, 2018 at 5:27
• @E.H.E Both other answers handwave that $\,(x^a)^b=x^{a \cdot b}\,$, which is not true in general. Even if that works in this case, any answer that doesn't emphasize why it happens to work here, and where else it does not work, is a dangerously incomplete answer. Whoever downvoted this answer did not understand that point, and is at risk of falling for any one of those $\,-1=(-1)^{2/2}=\sqrt{(-1)^2}=1\,$ perennial questions that get posted on MSE every other week.
– dxiv
Commented Jan 21, 2018 at 5:40