Question:
Find the value
$a^{12}+b^{12}$
Solution:
$=\left(a^{4}\right)^{3}+\left(b^{4}\right)^{3}$
$=\left(a^{4}+b^{4}\right)\left(\left(a^{4}\right)^{2}-a^{4} \times b^{4}+\left(b^{4}\right)^{2}\right)$
$\therefore\left[a^{3}+b^{3}=(a+b)\left(a^{2}-a b+b^{2}\right)\right]$
$=\left(a^{4}+b^{4}\right)\left(a^{8}-a^{4} b^{4}+b^{8}\right)$
$\therefore a^{12}+b^{12}=\left(a^{4}+b^{4}\right)\left(a^{8}-a^{4} b^{4}+b^{8}\right)$
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