Question

$3\sqrt{3}{x}^3+y^3$ is equal to

Answer 1

Given, $3\sqrt{3}{x}^3+y^3=(\sqrt{3}x{)}^3+y^3$
Since, $x^3+y^3=(x+y)(x^2+y^2-xy)$
$\therefore (\sqrt{3}x{)}^3+y^3=(\sqrt{3}x+y\left)\right((\sqrt{3}x{)}^2+y^2-\sqrt{3}xy)$
$=(\sqrt{3}x+y)(3x^2+y^2-\sqrt{3}xy)$
$=(\sqrt{3}x+y)(3x^2-\sqrt{3}xy+y^2)$
Option A is correct.