Question:
Find the value
$32 a^{3}+108 b^{3}$
Solution:
$=4\left(8 a^{3}+27 b^{3}\right)$
$=4\left((2 a)^{3}+(3 b)^{3}\right)$
$=4\left[(2 a+3 b)\left((2 a)^{2}-2 a \times 3 b+(3 b)^{2}\right.\right.$
$\therefore\left[a^{3}+b^{3}=(a+b)\left(a^{2}-a b+b^{2}\right)\right]$
$=4(2 a+3 b)\left(4 a^{2}-6 a b+9 b^{2}\right)$
$\therefore 32 a^{3}+108 b^{3}=4(2 a+3 b)\left(4 a^{2}-6 a b+9 b^{2}\right)$
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