Solve [(1/4x+3/2y)(1/4x-3/2y)-(frac{1}{4}x-frac{3}{2}y)^{2}+frac{9}{2}y^2]^2(4x^2-8xy):(3xy)^2=

$$[ ( \frac { 1 } { 4 } x + \frac { 3 } { 2 } y ) ( \frac { 1 } { 4 } x - \frac { 3 } { 2 } y ) - ( \frac { 1 } { 4 } x - \frac { 3 } { 2 } y ) ^ { 2 } + \frac { 9 } { 2 } y ^ { 2 } ] ^ { 2 } ( 4 x ^ { 2 } - 8 x y ) : ( 3 x y ) ^ { 2 } =$$

Evaluate

$$\frac{x\left(x-2y\right)}{4}$$

Short Solution Steps

Expand \left(3xy\right)^{2}.

$$\frac{\left(\left(\frac{1}{4}x+\frac{3}{2}y\right)\left(\frac{1}{4}x-\frac{3}{2}y\right)-\left(\frac{1}{4}x-\frac{3}{2}y\right)^{2}+\frac{9}{2}y^{2}\right)^{2}\left(4x^{2}-8xy\right)}{3^{2}x^{2}y^{2}}$$

Calculate 3 to the power of 2 and get 9.

$$\frac{\left(\left(\frac{1}{4}x+\frac{3}{2}y\right)\left(\frac{1}{4}x-\frac{3}{2}y\right)-\left(\frac{1}{4}x-\frac{3}{2}y\right)^{2}+\frac{9}{2}y^{2}\right)^{2}\left(4x^{2}-8xy\right)}{9x^{2}y^{2}}$$

Factor the expressions that are not already factored.

$$\frac{4x\left(x-2y\right)\left(\left(\frac{1}{4}x-\frac{3}{2}y\right)\left(\frac{1}{4}x+\frac{3}{2}y\right)-\left(\frac{1}{4}x-\frac{3}{2}y\right)^{2}+\frac{9}{2}y^{2}\right)^{2}}{9x^{2}y^{2}}$$

Cancel out x in both numerator and denominator.

$$\frac{4\left(x-2y\right)\left(\left(\frac{1}{4}x-\frac{3}{2}y\right)\left(\frac{1}{4}x+\frac{3}{2}y\right)-\left(\frac{1}{4}x-\frac{3}{2}y\right)^{2}+\frac{9}{2}y^{2}\right)^{2}}{9xy^{2}}$$

Expand the expression.

$$\frac{-\frac{9}{2}x^{2}y^{3}+\frac{9}{4}y^{2}x^{3}}{9xy^{2}}$$

Factor the expressions that are not already factored.

$$\frac{\frac{9}{4}\left(x-2y\right)x^{2}y^{2}}{9xy^{2}}$$

Cancel out xy^{2} in both numerator and denominator.

$$\frac{\frac{9}{4}x\left(x-2y\right)}{9}$$

Expand the expression.

$$\frac{\frac{9}{4}x^{2}-\frac{9}{2}xy}{9}$$

Expand

$$-\frac{xy}{2}+\frac{x^{2}}{4}$$

Short Solution Steps

Expand \left(3xy\right)^{2}.

$$\frac{\left(\left(\frac{1}{4}x+\frac{3}{2}y\right)\left(\frac{1}{4}x-\frac{3}{2}y\right)-\left(\frac{1}{4}x-\frac{3}{2}y\right)^{2}+\frac{9}{2}y^{2}\right)^{2}\left(4x^{2}-8xy\right)}{3^{2}x^{2}y^{2}}$$

Calculate 3 to the power of 2 and get 9.

$$\frac{\left(\left(\frac{1}{4}x+\frac{3}{2}y\right)\left(\frac{1}{4}x-\frac{3}{2}y\right)-\left(\frac{1}{4}x-\frac{3}{2}y\right)^{2}+\frac{9}{2}y^{2}\right)^{2}\left(4x^{2}-8xy\right)}{9x^{2}y^{2}}$$

Factor the expressions that are not already factored.

$$\frac{4x\left(x-2y\right)\left(\left(\frac{1}{4}x-\frac{3}{2}y\right)\left(\frac{1}{4}x+\frac{3}{2}y\right)-\left(\frac{1}{4}x-\frac{3}{2}y\right)^{2}+\frac{9}{2}y^{2}\right)^{2}}{9x^{2}y^{2}}$$

Cancel out x in both numerator and denominator.

$$\frac{4\left(x-2y\right)\left(\left(\frac{1}{4}x-\frac{3}{2}y\right)\left(\frac{1}{4}x+\frac{3}{2}y\right)-\left(\frac{1}{4}x-\frac{3}{2}y\right)^{2}+\frac{9}{2}y^{2}\right)^{2}}{9xy^{2}}$$

Expand the expression.

$$\frac{-\frac{9}{2}x^{2}y^{3}+\frac{9}{4}y^{2}x^{3}}{9xy^{2}}$$

Factor the expressions that are not already factored.

$$\frac{\frac{9}{4}\left(x-2y\right)x^{2}y^{2}}{9xy^{2}}$$

Cancel out xy^{2} in both numerator and denominator.

$$\frac{\frac{9}{4}x\left(x-2y\right)}{9}$$

Expand the expression.

$$\frac{\frac{9}{4}x^{2}-\frac{9}{2}xy}{9}$$