Question
Factorise:
$a^{12}-b^{12}$
$a^{12}-b^{12}$
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Answer
$a^{12}-b^{12}$
$=\left(a^6\right)^2-\left(b^6\right)^2$
$=\left(a^6-b^6\right)\left(a^6+b^6\right)$
$=\left[\left(a^3\right)^2-\left(b^3\right)^2\right]\left[\left(a^2\right)^3+\left(b^2\right)^3\right]$
$=\left(a^3-b^3\right)\left(a^3+b^3\right)\left[\left(a^2+b^2\right)\left(a^4-a^2 b^2+b^4\right)\right]$
$=(a-b)\left(a^2+a b+b^2\right)(a+b)\left(a^2-a b+b^2\right)\left(a^2+b^2\right)\left(a^4-a^2 b^2+b^4\right)$
$=(a-b)(a+b)\left(a^2+b^2\right)\left(a^2+a b+b^2\right)\left(a^2-a b+b^2\right)\left(a^4-a^2 b^2+b^4\right)$
$=\left(a^6\right)^2-\left(b^6\right)^2$
$=\left(a^6-b^6\right)\left(a^6+b^6\right)$
$=\left[\left(a^3\right)^2-\left(b^3\right)^2\right]\left[\left(a^2\right)^3+\left(b^2\right)^3\right]$
$=\left(a^3-b^3\right)\left(a^3+b^3\right)\left[\left(a^2+b^2\right)\left(a^4-a^2 b^2+b^4\right)\right]$
$=(a-b)\left(a^2+a b+b^2\right)(a+b)\left(a^2-a b+b^2\right)\left(a^2+b^2\right)\left(a^4-a^2 b^2+b^4\right)$
$=(a-b)(a+b)\left(a^2+b^2\right)\left(a^2+a b+b^2\right)\left(a^2-a b+b^2\right)\left(a^4-a^2 b^2+b^4\right)$
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