1. ## factorizing

factorize $\displaystyle (x+3y)^3-x^3-27y^3$

2. Originally Posted by requal
factorize $\displaystyle (x+3y)^3-x^3-27y^3$
I suggest that you first expand the perfect cube and then simplify the overall expression. You will be left with two terms, which you can factorise by taking out a common factor.

3. $\displaystyle (x+3y)^3-x^3-27y^3=9 x y (x+3 y)$

4. Hello,

Use this identity :
$\displaystyle a^3+b^3=(a+b)(a^2-ab+b^2)$

We here have :
$\displaystyle (x+3y)^3-(x^3+27y^3)=(x+3y)^3-(x^3+(3y)^3)=(x+3y)^3-(x+3y)(x^2-3xy+9y^2)$

$\displaystyle =(x+3y) \left((x+3y)^2-(x^2-3xy+9y^2)\right)$

$\displaystyle =(x+3y) (x^2+6xy+9y^2-x^2+3xy-9y^2)$

$\displaystyle =(x+3y)(9xy)$
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