How can I simplify the derivative further to match the correct answer? I've been stuck on trying to simplify my expression in order to match the correct answer but I can't seem to get the correct solution. Guidance towards the proper steps would be greatly appreciated!
The original question:

Differentiate and simplify completely
  $$g(x)=\sqrt[3]{\frac{x^3-1}{x^3+1}}$$

My solution:

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

I need help simplifying this to become this solution:

$$g^\prime(x) = \frac{2x^2\;\sqrt[3]{\left(x^3+1\right)^2}\;\sqrt[3]{x^3-1}}{(x+1)^2\left(x^2-x+1\right)^2(x-1)\left(x^2+x+1\right)} $$

Original image of solution.
 A: To remove some visual clutter, I'll define $a:=x^3-1$ and $b:=x^3+1$. So, your solution looks like:
$$\frac{2x^2}{\sqrt[3]{a^2}\;\sqrt[3]{b^4}} \tag{1}$$
Note that $\sqrt[3]{b^4}=\sqrt[3]{b^3 b} = b\sqrt[3]{b}$, so that $(1)$ becomes
$$\frac{2x^2}{b\;\sqrt[3]{a^2}\;\sqrt[3]{b}}\tag{2}$$
From here, we want to rationalize the denominator. We'll multiply that denominator by things that eliminate the cube roots, but then multiply the numerator by the same things for balance. Since
$$\sqrt[3]{w^2}\cdot\sqrt[3]{w}= \sqrt[3]{w^3} = w \tag{3}$$
multiplying by $\sqrt[3]{a}$ and $\sqrt[3]{b^2}$ will do what we want; thus,
$$\frac{2x^2}{b\;\sqrt[3]{a^2}\;\sqrt[3]{b}}\cdot\frac{\sqrt[3]{a}}{\sqrt[3]{a}}\cdot\frac{\sqrt[3]{b^2}}{\sqrt[3]{b^2}}=\frac{2x^2\;\sqrt[3]{a}\;\sqrt[3]{b^2}}{b\cdot ab} = \frac{2x^2\;\sqrt[3]{a}\;\sqrt[3]{b^2}}{ab^2}\tag{4}$$
That's essentially it, except for replacing $a$ and $b$ with their $x$-forms.
However, the target solution factors those $x$-forms (although, curiously, only in the denominator; go figure), using identities that you may want to commit to memory:
$$x^3-1=(x-1)(x^2+x+1) \qquad\qquad x^3+1=(x+1)(x^2-x+1) \tag{5}$$
This gives the target form, which I won't bother to type-out again. $\square$
A: Just multiply by $1$ (to free the denominator from irrationality):
$$g^\prime(x) = \frac{2x^2}{\sqrt[3]{\left(x^3-1\right)^2}\sqrt[3]{\left(x^3+1\right)^4}}\cdot \frac{\sqrt[3]{\left(x^3+1\right)^2}\;\sqrt[3]{x^3-1}}{\sqrt[3]{\left(x^3+1\right)^2}\;\sqrt[3]{x^3-1}}=\\
\frac{2x^2\;\sqrt[3]{\left(x^3+1\right)^2}\;\sqrt[3]{x^3-1}}{\color{red}{(x^3+1)^2}\color{blue}{(x^3-1)}}=\\
\frac{2x^2\;\sqrt[3]{\left(x^3+1\right)^2}\;\sqrt[3]{x^3-1}}{\color{red}{(x+1)^2\left(x^2-x+1\right)^2}\color{blue}{(x-1)\left(x^2+x+1\right)}}.$$
