Question
Simplify using following identity :$(a \pm b)\left(a^2 \pm a b+b^2\right)=a^3 \pm b^3 \left(\frac{a}{3}-3 b\right)\left(\frac{a^2}{9}+a b+9 b^2\right)$

Answer

$ \left(\frac{a}{3}-3 b\right)\left(\frac{a^2}{9}+a b+9 b^2\right)$
$ =\left(\frac{a}{3}-3 b\right)\left[\left(\frac{a}{3}\right)^2+\left(\frac{a}{3}\right)(3 b)+(3 b)^2\right]$
$ =\left(\frac{a}{3}\right)^3-(3 b)^3$
$ =\frac{a^3}{27}-27 b^3$

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