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...
 A: One way to go is to use $a^3-b^3=(a-b)(a^2+ab+b^2)$.
So with $a=1$ and $b=\sqrt[3]{t}$:
$$\frac{1-t}{1-\sqrt[3]{t}} = \frac{1-(\sqrt[3]{t})^3}{1-\sqrt[3]{t}}=1^2+1.\sqrt[3]{t}+(\sqrt[3]{t})^2$$
And with $a=1$ and $b=-\sqrt[3]{t}$:
$$\frac{1+t}{1+\sqrt[3]{t}} = \frac{1-(-\sqrt[3]{t})^3}{1+\sqrt[3]{t}}=1^2-1.\sqrt[3]{t}+(-\sqrt[3]{t})^2$$
So $$A(t)=2 +2(\sqrt[3]{t})^2$$
A: 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.
A: By using the substitution $\sqrt[3]t=a$, we have, $$\dfrac{1-t}{1-\sqrt[3]{t}}+\dfrac{1+t}{1+\sqrt[3]{t}}=\dfrac{a^3-1}{a-1}+\dfrac{a^3+1}{a+1}=(a^2+a+1)+(a^2-a+1)=2a^2+2$$
Which is equal to $2+2\sqrt[3]{t^2}$.
