I need help with the following problem:
Rewrite the following expression so that the denominator contains no root expression and the fraction has been shortened as far as possible.
I have tried to express it as
But I'm not sure how to continue from there.
When we have a power of a power we can multiply the exponents, this is to say that for three numbers $a, m, n$ it is true that $$ (a^m)^n = a^{mn}$$.
Notice that $8 = 2^3$ and $64 = 4^3$. We can simplify this a bit by
$$ y = \frac{\sqrt[3]{x} + \sqrt[3]{8}}{\sqrt[3]{64} - \sqrt[3]{8x} + \sqrt[3]{x^2}} = \frac{\sqrt[3]{x} + \sqrt[3]{2^3}}{\sqrt[3]{4^3} - \sqrt[3]{2^3 \cdot x} + \sqrt[3]{x^2}} = \frac{\sqrt[3]{x} + 2}{4 - 2 \sqrt[3]{x} + \sqrt[3]{x^2}} $$
And we can factor a sum of two cubes as
$$ a^3 + b^3 = (a+b)(a^2 - ab + b^2) $$ (Please verify this in your own.) It is a little tricky to notice at first, but it is what you have in the bottom of the fraction. It may be easier to see if we set $u = \sqrt[3]{x}$
$$ y = \frac{u + 2}{4 - 2u + u^2} = \frac{u + 2}{u^2 - 2u + 4} = \frac{u + 2}{u^2 - 2u + 2^2} $$
Now, multiply both sides by $\frac{u+2}{u+2}=1$ (we can do this becouse multiply by 1 does nothing), and get
$$ y = \frac{(u + 2)(u + 2)}{(u + 2)(u^2 - 2u + 2^2)} = \frac{u^2 + 4u + 4}{u^3 + 2^3} = \frac{u^2 + 4u + 4}{u^3 + 8} $$
And just substitute that $u = \sqrt[3]{x}$, and we have
$$ y = \frac{(\sqrt[3]{x})^2 + 4\sqrt[3]{x} + 4}{(\sqrt[3]{x})^3 + 8} = \frac{\sqrt[3]{x^2} + 4\sqrt[3]{x} + 4}{x + 8} $$