Question

Factorise $3\sqrt{3}{x}^3+y^3$

• A. $(\sqrt{3}x+y)(3x^2-\sqrt{3}xy+y^2)$
• B. $(\sqrt{3}x-y)(\sqrt{3}{x}^2-\sqrt{3}xy+y^2)$
• C. $(\sqrt{3}x+y)(3x^2+\sqrt{3}xy+y^2)$
• D. $(\sqrt{3}x-y)(3x^2-\sqrt{3}xy-y^2)$

Answer 1

The correct option is A $(\sqrt{3}x+y)(3x^2-\sqrt{3}xy+y^2)$

As we know that,

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

Therefore,

$3\sqrt{3}{x}^3+y^3$

$={\left(\sqrt{3}x\right)}^3+y^3$

$=(\sqrt{3}x+y)({\left(\sqrt{3}x\right)}^2-\left(\sqrt{3}x\right)y+y^2)$

$=(\sqrt{3}x+y)(3x^2-\sqrt{3}xy+y^2)$