Question
Find the cube of$: 3 a+\frac{1}{3 a}$
✓
Answer
$\left(3 a+\frac{1}{3 a}\right)^3 $
$=(3 a)^3+\left(\frac{1}{3 a}\right)^3+3(3 a)\left(\frac{1}{3 a}\right)\left(3 a+\frac{1}{3 a}\right) $
$=27 a^3+\frac{1}{27 a^3}+9 a+\frac{1}{a} .$
$=(3 a)^3+\left(\frac{1}{3 a}\right)^3+3(3 a)\left(\frac{1}{3 a}\right)\left(3 a+\frac{1}{3 a}\right) $
$=27 a^3+\frac{1}{27 a^3}+9 a+\frac{1}{a} .$
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