Determine the value without solving for a I got this question at the bottom of this page, the last question on the page. I've been taught how to handle this question but I couldn't figure this one out so I'd like to show how the website solved it and I please want someone to explain to me the steps they took to get to their answer. Assume that I know nothing about algebra. 
Given that $a+\frac{1}{3a}=2$, determine the value of $a^3+\frac{1}{27a^3}$ without solving for $a$.
$$a^3+\frac{1}{27a^3} = (a+\frac{1}{3a})(a^2-\frac{1}{3}+\frac{1}{9a^2})=2(a^2-\frac{1}{3}+\frac{1}{9a^2})$$
$$a^2-\frac{1}{3}+\frac{1}{9a^2}=(a+\frac{1}{3a})^2-1=4-1=3$$
$$a^3+\frac{1}{27a^3}=2(3)=6$$
Apologies, I don't know how to put new lines in the middle of equations so if you want clarity you can check on the site, it's literally the last question on the page. 

hint
use

$$(a+b)^3=a^3+3ab (a+b)+b^3$$

with $b=\frac {1}{3a} $, observe that
$3ab=1$,
and $$27=3^3$$
With $a+b=2$, the result is $$2^3-2=6$$
