Question

Factorise $27a^3+b^3+8c^3-18abc$ using identity

Answer 1

We have $27a^3+b^3+8c^3-18abc$ $=(3a{)}^3+(b{)}^3+(2c{)}^3-3(3a\left)\right(b\left)\right(2c)$

We know that $x^3+y^3+z^3-3xyz=(x+y+z)(x^2+y^2+z^2-xy-yz-zx)$

Comparing both, we get

$x=3a,y=b$ and $z=2c$

Then $(3a+b+2c)\left[\right(3a{)}^2+(b{)}^2+(2c{)}^2-\left(3a\right)\left(b\right)-\left(b\right)\left(2c\right)-\left(2c\right)\left(3a\right)]$

$=(3a+b+2c)(9a^2+b^2+4c^2-3ab-2bc-6ac)$