To solve this problem, you have to know these two special factorizations:
[tex] x^3-y^3=(x-y)(x^2+xy+y^2)\\
x^3+y^3=(x+y)(x^2-xy+y^2) [/tex]
Knowing these tells us that if we want to rationalize the numerator. we want to use the top equation to our advantage. Let:
[tex] \sqrt[3]{x+h}=x\\
\sqrt[3]{x}=y [/tex]
That tells us that we have:
[tex] \frac{x-y}{h}[/tex]
So, since we have one part of the special factorization, we need to multiply the top and the bottom by the other part, so:
[tex] \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)} [/tex]
So, we have:
[tex] \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}} [/tex]
That is our rational expression with a rationalized numerator.
Also, you could just mutiply by:
[tex] \frac{1}{\sqrt[3]{x_h}-\sqrt[3]{x}} \text{ to get}\\
\frac{1}{h\sqrt[3]{x+h}-h\sqrt[3]{h}} [/tex]
Either way, our expression is rationalized.
