Question

Rationalize the numerator or denominator, and simplify:

Answer 1

$\dfrac{\sqrt[3]{x+h}-\sqrt[3]x}h\times\dfrac{\sqrt[3]{(x+h)^2}+\sqrt[3]{x(x+h)}+\sqrt[3]{x^2}}{\sqrt[3]{(x+h)^2}+\sqrt[3]{x(x+h)}+\sqrt[3]{x^2}}=\dfrac{(\sqrt[3]{x+h})^3-(\sqrt[3]x)^3}\cdots$
$=\dfrac{x+h-x}\cdots=\dfrac h\cdots$

The $h$s then cancel, leaving you with the $\cdots=\sqrt[3]{(x+h)^2}+\sqrt[3]{x(x+h)}+\sqrt[3]{x^2}$ term.

If it's not clear what I did above, consider the substitution $a=\sqrt[3]{x+h}$ and $b=\sqrt[3]x$. Then

$a^3-b^3=(a-b)(a^2+ab+b^2)$
$\implies\dfrac{a-b}h=\dfrac{a-b}h\times\dfrac{a^2+ab+b^2}{a^2+ab+b^2}=\dfrac{a^3-b^3}{h(a^2+ab+b^2)}$