Simplify $A(t)=\frac{1-t}{1-\sqrt[3]{t}}+\frac{1+t}{1+\sqrt[3]{t}}$

Solution 1:

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 $$A(t)=2 +2(\sqrt[3]{t})^2$$

Solution 2:

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.

Solution 3:

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 $2+2\sqrt[3]{t^2}$.