Question

Factorise:
$81x^3-3y^3$

• A. $3(3x-y)[9x^2+3xy+y^2]$
• B. $3(9x-y)[9x^2+3xy+y^2]$
• C. $(3x-y)[9x^2+3xy+y^2]$
• D. $3(3x-y)[9x^2+6xy+y^2]$

Answer 1

The correct option is A $3(3x-y)[9x^2+3xy+y^2]$

$81x^3-3y^3=3(27x^3-y^3)$
$=3(3^3{x}^3-y^3)$
$=3\left[\right(3x{)}^3-y^3]$
Comparing this with $a^3-b^3,$
we get a = 3x, b = y
Using the identity,
$a^3-b^3=(a-b)(a^2+ab+b^2)$
$⟹3(3x-y)\left[\right(3x{)}^2+\left(3x\right)\left(y\right)+y^2]$
$\therefore 81x^3-3y^3=3(3x-y)[9x^2+3xy+y^2]$