$\sqrt[4]{(-x)^4}=x$; Is the condition true always, sometimes, or never?

Question

$$\sqrt[4]{(-x)^4}=x$$

This equation was provided to me, and I had to find out if this condition was sometimes, always, or never true.

At first my initial thought was that the condition will ALWAYS be true, since $\sqrt[4]{(-x)^4}$ evaluates to $\sqrt[4]{x^4}$, which evaluates to $x$. Since $x=x$, the condition is always true.

However, as I was about to put Always as my solution, I realized that SOMETIMES might be the correct answer. My process was that $\sqrt[4]{(-x)^4}$ can be restated as $(-x)^{4^{\frac{1}{4}}}$, and after multiplying the exponents, this is simply $(-x)^1$ or $-x$.

$-x=x$ is only true for the value $x=0$, so I put SOMETIMES as my solution.

Is sometimes, or always the solution to this question?

If you find the solution to this, remember, you are answering this important question:

In which order do you simplify the expression, do you carry out the exponent to the base first, or do you simplify exponents first?

Answer 1

It is always true that $$\sqrt[4]{(-x)^4}=|x|$$

Indeed by definition the square root of $a\geq0$ $$\sqrt[4] a$$ is the non negative value $b$ such that $$b^4=a$$

Answer 2

It should be sometimes since there are two cases to consider:

CASE 1 ($x \ge 0$): In this case, we have $\sqrt[4]{(-x)^4}=\sqrt[4]{x^4} = x$

CASE 2 ($x < 0$): In this case, we have $\sqrt[4]{(-x)^4} = \sqrt[4]{x^4} = -x$ because $x$ is already a negative number and the result should be positive.

Here, notice that we always do the operation $(-x)^4 = x^4$ first, then the $\sqrt[4]{x^4} = x$ or $-x$ according to the value of $x$.

And we can combine these results as $\sqrt[4]{(-x)^{4}} = |x|$ as you can see from the above cases.