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Let $x$ be a real number. Consider the following expression:

$$ \sqrt[3]{\frac{x^{3} - 3x + \left(x^{2} - 1\right)\sqrt{x^{2} - 4}}{2}} + \sqrt[3]{\frac{x^{3} - 3x - \left(x^{2} - 1\right)\sqrt{x^{2} - 4}}{2}} $$

It includes an arithmetic square root, so it's defined only if the expression under that root is non-negative, namely:

$$ x^{2} - 4 \geq 0 \Longleftrightarrow x^{2} \geq 4 \Longleftrightarrow |x| \geq 2 \Longleftrightarrow \left[ \begin{array}[l]\ x \geq 2, \\ x \leq -2. \end{array} \right. $$

We will now simplify it. Here is the approach I came up with.

Define the following:

\begin{align} A &:= \sqrt[3]{\frac{x^{3} - 3x + \left(x^{2} - 1\right)\sqrt{x^{2} - 4}}{2}}, \\ B &:= \sqrt[3]{\frac{x^{3} - 3x - \left(x^{2} - 1\right)\sqrt{x^{2} - 4}}{2}}, \\ u &:= A + B. \end{align}

After cubing both sides of the last expression, we get:

\begin{align} u^{3} &= \left(A + B\right)^{3}, \\ u^{3} &= A^{3} + B^{3} + 3AB\left(A + B\right), \\ u^{3} &= A^{3} + B^{3} + 3ABu. \end{align}

Now, let's find out the value of $A^{3} + B^{3}$:

\begin{align}\require{cancel} A^{3} + B^{3} &= \left(\sqrt[3]{\frac{x^{3} - 3x + \left(x^{2} - 1\right)\sqrt{x^{2} - 4}}{2}}\right)^{3} + \left(\sqrt[3]{\frac{x^{3} - 3x - \left(x^{2} - 1\right)\sqrt{x^{2} - 4}}{2}}\right)^{3} \\ &= \frac{x^{3} - 3x + \left(x^{2} - 1\right)\sqrt{x^{2} - 4}}{2} + \frac{x^{3} - 3x - \left(x^{2} - 1\right)\sqrt{x^{2} - 4}}{2} \\ &= \frac{1}{2}\left(x^{3} - 3x \cancel{+ \left(x^{2} - 1\right)\sqrt{x^{2} - 4}} + x^{3} - 3x \cancel{- \left(x^{2} - 1\right)\sqrt{x^{2} - 4}}\right) \\ &= \frac{1}{2}\left((x^{3} - 3x) + (x^{3} - 3x)\right) \\ &= \frac{1}{2}\cdot 2(x^{3} - 3x) \\ &= x^{3} - 3x \end{align}

Before working with $AB$, note the following:

\begin{align} \left(x^{2} - 1\right)\sqrt{x^{2} - 4} &= \sqrt{\left(x^{2} - 1\right)^{2}\left(x^{2} - 4\right)} \\ &= \sqrt{\left(x^{4} - 2x^{2} + 1\right)\left(x^{2} - 4\right)} \\ &= \sqrt{x^{6} - 6x^{4} + 9x^{2} - 4} \\ &= \sqrt{\left(x^{3}\right)^{2} - 2x^{3}\cdot 3x + \left(3x\right)^{2} - 4} \\ &= \sqrt{\left(x^{3} - 3x\right)^{2} - 4} \end{align}

Now, let's find out the value of $AB$:

\begin{align}\require{cancel} AB &= \sqrt[3]{\frac{x^{3} - 3x + \left(x^{2} - 1\right)\sqrt{x^{2} - 4}}{2}} \cdot \sqrt[3]{\frac{x^{3} - 3x - \left(x^{2} - 1\right)\sqrt{x^{2} - 4}}{2}} \\ &= \sqrt[3]{\frac{1}{2}\left(x^{3} - 3x + \left(x^{2} - 1\right)\sqrt{x^{2} - 4}\right)\cdot \frac{1}{2}\left(x^{3} - 3x - \left(x^{2} - 1\right)\sqrt{x^{2} - 4}\right)} \\ &= \sqrt[3]{\frac{1}{4}\left((x^{3} - 3x) + \left(x^{2} - 1\right)\sqrt{x^{2} - 4}\right)\left((x^{3} - 3x) - \left(x^{2} - 1\right)\sqrt{x^{2} - 4}\right)} \\ &= \sqrt[3]{\frac{1}{4}\left((x^{3} - 3x)^2 - \left(x^{2} - 1\right)^2\left(x^{2} - 4\right)\right)} \\ &= \sqrt[3]{\frac{1}{4}\left((x^{3} - 3x)^2 - \left(\left(x^{3} - 3x\right)^{2} - 4\right)\right)} \\ &= \sqrt[3]{\frac{1}{4}\left(\cancel{(x^{3} - 3x)^2} \cancel{- \left(x^{3} - 3x\right)^{2}} + 4\right)} \\ &= 1 \end{align}

By substituting $A^{3}+B^{3}$ and $AB$ with their values, we have:

\begin{align} u^{3} &= A^{3} + B^{3} + 3ABu, \\ u^{3} &= x^{3} - 3x + 3u, \\ u^{3} - 3u &= x^{3} - 3x. \end{align}

Further on, we are going to use a couple of known facts:

  • A cubic $ax^{3} + bx^{2} + cx + d$ with $a, b, c, d \in \mathbb{R}$ has two local extrema if and only if $b^{2} > 3ac$, and is monotonic otherwise.
  • If a cubic has two local extrema, then the local maximum $x_{\max}$ can be found by the formula $x_{\max} = \frac{-b - \sqrt{b^{2} - 3ac}}{3a}$, and the local minimum $x_{\min}$ can be found by the formula $x_{\min} = \frac{-b + \sqrt{b^{2} - 3ac}}{3a}$.

In our case $b = 0$, and this simplifies our reasoning, as:

  • For checking for the local extrema, it's enough to check whether $-3ac > 0$.
  • The local maximum is found by $x_{\max} = - \frac{\sqrt{-3ac}}{3a}$, and the local minimum is found by $x_{\min} = \frac{\sqrt{-3ac}}{3a}$.

Consider the function $f(x) = x^{3} - 3x$. As $(-3)\cdot 1\cdot (-3) = 9 > 0$, it has two local extrema at:

$$ x_{\max} = -\frac{\sqrt{9}}{3} = -1,\ \ \ \ \ \ \ \ \ \ \ x_{\min} = \frac{\sqrt{9}}{3} = 1. $$

Therefore, $f(x)$ is decreasing on $(-1; 1)$ and increasing elsewhere. The expression in question is defined on $(-\infty; -2]\cup [2;+\infty)$. There, $f(x)$ is increasing. Thus, $u^{3} - 3u = x^{3} - 3x \Leftrightarrow u = x$ on that interval, so the answer is $x$.

Now, on to the question.

I'm sure there is a simpler approach to the same simplification.

I thought about completing the cube inside each cube root in order to cancel both roots with exponents and simplify what's left.

However, I can't seem to deduce a procedure to do this in such expressions. Is it possible, and if yes, then what's the general approach?

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  • $\begingroup$ The function is also defined in the real domain if $x=1+\epsilon$ $\endgroup$ Commented Nov 14 at 7:27
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    $\begingroup$ With hindsight, one sees that $(x+\sqrt{x^2-4})^3=4(x^3-3x+(x^2-1)\sqrt{x^2-4})$. $\endgroup$ Commented Nov 14 at 9:00
  • $\begingroup$ I doubt if there is any general procedure. Essentially, you will end up finding $u$ by solving a cubic - in this case, $u^3-3u-(x^3-3x)=0$, which is easy. In general, however, you will need one of the methods for solving a not-nice cubic: the algebraic ones lead to solutions in the form of the sum of two cube roots, which takes you back to where you started; my solution below uses another method, but it only works to give anything other than $\cosh(\frac{1}{3}\cosh^{-1}(...))$ because there is a neat answer. $\endgroup$ Commented Nov 14 at 10:40
  • $\begingroup$ If you want to simplify the last part of your argument, you could use the fact that $0 = u^3-3u-(x^3-3x)=(u-x)(u^2+ux +x^2 -3)=(u-x)\big((u+\frac{1}{2}x)^2+\frac{3}{4}(x^2-4)\big)$, which, for $x^2>4$, only has one real root (and $u$ is definitely real). $\endgroup$ Commented Nov 14 at 12:13

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For $x>0$, let $x=2\cosh \theta$ (which is always possible since $x\ge 2$). Then, using the triple angle identities for the hyperbolic functions, the content of the first cube root is $e^{3\theta}$ and of the second, $e^{-3\theta}$. Thus the sum of the cube roots is $2\cosh \theta=x$. The same argument works for $x<0$, with $x=-2\cosh \theta$.

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