Question

Factorise:

$9a^2+\frac{1}{9a^2}-2-12a+\frac{4}{3a}$

Answer 1

$9a^2+\frac{1}{9a^2}-2-12a+\frac{4}{3a}=\left[\right(3a{)}^2+(\frac{1}{3a}{)}^2-2\times (3a)\times (\frac{1}{3a}\left)\right]-4(3a-\frac{1}{3a})=(3a-\frac{1}{3a}{)}^2-4(3a-\frac{1}{3a})=(3a-\frac{1}{3a}\left)\right(3a-\frac{1}{3a}-4)$