Question

Factorise: $7y^3+12z^3$

Answer 1

$7y^3+12z^3=P$

$P=(\sqrt[3]{37}y{)}^3+(\sqrt[3]{312}z{)}^3$

We know $a^3+b^3=(a+b)(a^2-ab+b^2)$

$P=(\sqrt[3]{37}y+\sqrt[3]{312}z)\left(\right(7{)}^{\frac{2}{3}}+(12{)}^{\frac{2}{3}}{z}^2÷(84{)}^{\frac{2}{3}}{y}^z)$

$P=\left[\right(7{)}^{\frac{1}{3}}y+\left(12{)}^{\frac{1}{3}}z\right]\left[\right(7{)}^{\frac{2}{3}}{y}^2+(12{)}^{\frac{2}{3}}{z}^2-(84{)}^{\frac{2}{3}}{y}^z]$