Question

Factorise the following: $27(x-1{)}^3+y^3$

• A. $(3x-y-3)(9x^2+18x+9-3xy+3y+y^2)$
• B. $(3x+y-3)(9x^2-18x+9-3xy+3y+y^2)$
• C. $(3x+y-3)(x^2-18x+9-3xy+3y+y^2)$
• D. $(3x-y-3)(x^2-18x+9-3xy+3y+y^2)$

Answer 1

Answer: (B)

$27(x-1{)}^3+y^3$
Let $(x-1)=a$
Therefore,
$27a^3+y^3$
$=(3a{)}^3+(y{)}^3$
Using,$a^3+b^3=(a+b)(a^2-ab+b^2)$
$=\left[\right(3a+y\left)\right]\left[\right(3a{)}^2-\left(3a\right)\left(y\right)+\left(y{)}^2\right]$
$=(3a+y)\left[\right(9a{)}^2-3ay+\left(y{)}^2\right]$
Resubstituting the value we get,
$\left[3\right(x-1)+y]\left[9\right(x-1{)}^2-3y(x-1)+y^2]$
$=(3x-3+y)\left[9\right(x^2-2x+1)-3xy+3y+y^2]$
$=(3x+y-3)(9x^2-18x+9-3xy+3y+y^2)$