Question

Factorise: (1) $x^3+3y^3$
(2) $8y^3-8x^2-9$

Answer 1

(1) we need to factorise $x^3+3y^3$

By using the identity, $a^3+b^3=(a+b)(a^2-ab+b^2)$

Now we can rewrite the expression as $x^3+3y^3=(x+\sqrt{3}y)(x^2-\sqrt{3}xy+\sqrt{3}{y}^2)$

Hence, factors are $(x+\sqrt{3}y),(x^2-\sqrt{3}xy+\sqrt{3}{y}^2)$

(2). $8y^3-8x^2-9$

It cannot be factorised.