Question

find the product x^3y^3[-3/2(x-y)]​

Answer 1

Given:

x^3y^3[-3/2(x-y)]​

To find:

find the product x^3y^3[-3/2(x-y)]​

Solution:

From given, we have,

x^3y^3[-3/2(x-y)]​

$x^3y^3\left(-\frac{3}{2}\left(x-y\right)\right)\\\\=\left(xy\right)^3\left(-\frac{3}{2}\left(x-y\right)\right)\\\\=-\left(xy\right)^3\frac{3}{2}\left(x-y\right)\\\\=-\left(x-y\right)\frac{3\left(xy\right)^3}{2}\\\\=-\frac{3x^3y^3}{2}\left(x-y\right)$

∴ x^3y^3[-3/2(x-y)]​ = -[3x³y³/2](x - y)

Answer 2

Given: The term x^3y^3 [-3/2(x-y)]​

To find: The product.

Solution:

• Now we have given the term: x^3y^3 [-3/2(x-y)]​

• First solving the inner bracket, we get:

                 x^3y^3 [ -3x/2 + 3y/2 ]​

• Now multiplying , we get:

                 ( x^3y^3 × -3x/2 ) + ( x^3y^3 × y/2 )

                 -3x^4y^3 / 2 + 3x^3y^4 / 2

Answer:

          So the product is -3x^4y^3 / 2 + 3x^3y^4 / 2