Question

The value of $(2x-3y{)}^3+(x+3y{)}^3+3(2x-3y{)}^2(x+3y)+3(2x-3y\left)\right(x+3y{)}^2$ is

• A. 0
• B. $27x^3+9y^3$
• C. $27x^3$

Answer 1

The correct option is C $27x^3$

Substitute $2x-3y=a$ and $(x+3y)=b$
$(2x-3y{)}^3+(x+3y{)}^3+3(2x-3y{)}^2(x+3y)+3(2x-3y\left)\right(x+3y{)}^2$
$=a^3+b^3+3a^{2}b+3a{b}^2$
Now, on using the identity:
$(a+b{)}^3=a^3+b^3+3a^{2}b+3a{b}^2$
$=(a+b{)}^3$
$=(2x-3y+x+3y{)}^3$
$=(3x{)}^3$
$=27x^3$