factorize $\displaystyle (x+3y)^3-x^3-27y^3$
Hello,
Use this identity :
$\displaystyle a^3+b^3=(a+b)(a^2-ab+b^2)$
We here have :
$\displaystyle (x+3y)^3-(x^3+27y^3)=(x+3y)^3-(x^3+(3y)^3)=(x+3y)^3-(x+3y)(x^2-3xy+9y^2)$
$\displaystyle =(x+3y) \left((x+3y)^2-(x^2-3xy+9y^2)\right)$
$\displaystyle =(x+3y) (x^2+6xy+9y^2-x^2+3xy-9y^2)$
$\displaystyle =(x+3y)(9xy)$