Question
Factorise : $27 x^3+y^3+z^3-9 x y z$.
✓
Answer
$27 x^3+y^3+z^3-9 x y z$
$(3 x)^3+(y)^3+(z)^3-3(3 x)(y)(z)$
$=(3 x+y+z)\left\{(3 x)^2+(y)^2+(z)^2-(3 x)(y)-(y)(z)-(z)(3 x)\right\}$
(Using Identity $\left.a^3+b^3+c^3-3 a b c=(a+b+c)\left(a^2+b^2+c^2-a b-b c-c a\right)\right)$
$=(3 x+y+z)\left(9 x^2+y^2+z^2-3 x y-y z-3 z x\right)$
$(3 x)^3+(y)^3+(z)^3-3(3 x)(y)(z)$
$=(3 x+y+z)\left\{(3 x)^2+(y)^2+(z)^2-(3 x)(y)-(y)(z)-(z)(3 x)\right\}$
(Using Identity $\left.a^3+b^3+c^3-3 a b c=(a+b+c)\left(a^2+b^2+c^2-a b-b c-c a\right)\right)$
$=(3 x+y+z)\left(9 x^2+y^2+z^2-3 x y-y z-3 z x\right)$
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