Question

( 3x^2 - 2y )^3 ( 2/3 x^2 - 1/2 y )^3

Answer 1

( 3x^2 - 2y )^3 ( 2/3 x^2 - 1/2 y )^3
=\(\left[\left(3x^2-2y\right)\left(\dfrac{2}{3}x^2-\dfrac{1}{2}y\right)\right]^3\)
\(=\left(2x^2-\dfrac{3}{2}x^2y-\dfrac{4}{3}x^2y+y\right)^3\)
\(=\left(2x^2-\dfrac{1}{6}x^2y+y\right)^3\)
 

Answer 2

\(\left(3x^2-2y\right)^3=27x^6-54x^4y+36x^2y^2-8y^3\)

\(\left(\dfrac{2}{3}x^2-\dfrac{1}{2}y\right)^3=\dfrac{8}{27}x^6-\dfrac{2}{3}x^4y+\dfrac{1}{2}x^2y^2-\dfrac{1}{8}y^3\)