Solve (2x-3y)*(2x+3y)-(2x-3y)^2+6*(3y^2-2xy)

$$( 2 x - 3 y ) \cdot ( 2 x + 3 y ) - ( 2 x - 3 y ) ^ { 2 } + 6 \cdot ( 3 y ^ { 2 } - 2 x y )$$

Evaluate

$$0$$

Solution Steps

Consider \left(2x-3y\right)\left(2x+3y\right). Multiplication can be transformed into difference of squares using the rule: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}.

$$\left(2x\right)^{2}-\left(3y\right)^{2}-\left(2x-3y\right)^{2}+6\left(3y^{2}-2xy\right)$$

Expand \left(2x\right)^{2}.

$$2^{2}x^{2}-\left(3y\right)^{2}-\left(2x-3y\right)^{2}+6\left(3y^{2}-2xy\right)$$

Calculate 2 to the power of 2 and get 4.

$$4x^{2}-\left(3y\right)^{2}-\left(2x-3y\right)^{2}+6\left(3y^{2}-2xy\right)$$

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

$$4x^{2}-3^{2}y^{2}-\left(2x-3y\right)^{2}+6\left(3y^{2}-2xy\right)$$

Calculate 3 to the power of 2 and get 9.

$$4x^{2}-9y^{2}-\left(2x-3y\right)^{2}+6\left(3y^{2}-2xy\right)$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(2x-3y\right)^{2}.

$$4x^{2}-9y^{2}-\left(4x^{2}-12xy+9y^{2}\right)+6\left(3y^{2}-2xy\right)$$

To find the opposite of 4x^{2}-12xy+9y^{2}, find the opposite of each term.

$$4x^{2}-9y^{2}-4x^{2}+12xy-9y^{2}+6\left(3y^{2}-2xy\right)$$

Combine 4x^{2} and -4x^{2} to get 0.

$$-9y^{2}+12xy-9y^{2}+6\left(3y^{2}-2xy\right)$$

Combine -9y^{2} and -9y^{2} to get -18y^{2}.

$$-18y^{2}+12xy+6\left(3y^{2}-2xy\right)$$

Use the distributive property to multiply 6 by 3y^{2}-2xy.

$$-18y^{2}+12xy+18y^{2}-12xy$$

Combine -18y^{2} and 18y^{2} to get 0.

$$12xy-12xy$$

Combine 12xy and -12xy to get 0.

$$0$$

Factor

$$0$$