<s>[INST] Simplify $A(t)=\frac{1-t}{1-\sqrt[3]{t}}+\frac{1+t}{1+\sqrt[3]{t}}$ Simplify $$A(t)=\dfrac{1-t}{1-\sqrt[3]{t}}+\dfrac{1+t}{1+\sqrt[3]{t}}$$ and calculate $A(3\sqrt3).$ For $t\ne\pm1$ we have, $$A=\dfrac{(1-t)(1+\sqrt[3]{t})+(1+t)(1-\sqrt[3]{t})}{1-\sqrt[3]{t^2}}=\\=\dfrac{2-2t\sqrt[3]{t}}{1-\sqrt[3]{t^2}}$$ What to do next? I can't come up with anything else...
 [/INST] As a general principle, sometimes things look simpler if we perform a suitable substitution:  let $u = \sqrt[3]{t}$, so that $t = u^3$ and $$A(t) = A(u^3) = \frac{1 - u^3}{1 - u} + \frac{1 + u^3}{1 + u}.$$  Now it becomes obvious that we either need to factor the numerators, or put everything over a common denominator.  In the first approach, the difference of cubes factorization
$$1 \pm u^3 = (1 \pm u)(1 \mp u + u^2)$$
yields
$$A(u^3) = (1 + u + u^2) + (1 - u + u^2) = 2(1 + u^2),$$
hence $$A(t) = 2(1 + t^{2/3}).$$
In the second case,
$$\begin{align}
A(u^3) &= \frac{(1-u^3)(1+u) + (1+u^3)(1-u)}{(1-u)(1+u)} \\
&= \frac{(1 + u - u^3 - u^4) + (1 - u + u^3 - u^4)}{1-u^2} \\
&= \frac{2(1-u^4)}{1-u^2} \\
&= \frac{2(1+u^2)(1-u^2)}{1-u^2} \\
&= 2(1+u^2),
\end{align}$$ which is the same as the first approach.  When dealing with rational powers, it is often easier to visualize and manipulate the expression if we use a carefully chosen substitution.
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