Simplify the cube root

Question

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.

Answer 1

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

Answer 2

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

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.

Answer 3

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 $x$)