Question

How do you rationalize the numerator in this problem?

Answer 1

To solve this problem, you have to know these two special factorizations:

$x^3-y^3=(x-y)(x^2+xy+y^2)\\ x^3+y^3=(x+y)(x^2-xy+y^2)$

Knowing these tells us that if we want to rationalize the numerator. we want to use the top equation to our advantage. Let:

$\sqrt[3]{x+h}=x\\ \sqrt[3]{x}=y$

That tells us that we have:

$\frac{x-y}{h}$

So, since we have one part of the special factorization, we need to multiply the top and the bottom by the other part, so:

$\frac{x-y}{h}*\frac{x^2+xy+y^2}{x^2+xy+y^2}=\frac{x^3-y^3}{h*(x^2+xy+y^2)}$

So, we have:

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

That is our rational expression with a rationalized numerator.

Also, you could just mutiply by:

$\frac{1}{\sqrt[3]{x_h}-\sqrt[3]{x}} \text{ to get}\\ \frac{1}{h\sqrt[3]{x+h}-h\sqrt[3]{h}}$

Either way, our expression is rationalized.