Solve {l}{(1/2a-b)^2-(1/2a+b)^2]}{-3/2x(4/9x-2y)-(x-2y)}

$$\left. \begin{array} { l } { ( \frac { 1 } { 2 } a - b ) ^ { 2 } - ( \frac { 1 } { 2 } a + b ) ^ { 2 } ] } \\ { - \frac { 3 } { 2 } x ( \frac { 4 } { 9 } x - 2 y ) - ( x - 2 y ) } \end{array} \right.$$

Evaluate

$$-2ab$$

Solution Steps

Use binomial theorem \left(p-q\right)^{2}=p^{2}-2pq+q^{2} to expand \left(\frac{1}{2}a-b\right)^{2}.

$$\frac{1}{4}a^{2}-ab+b^{2}-\left(\frac{1}{2}a+b\right)^{2}$$

Use binomial theorem \left(p+q\right)^{2}=p^{2}+2pq+q^{2} to expand \left(\frac{1}{2}a+b\right)^{2}.

$$\frac{1}{4}a^{2}-ab+b^{2}-\left(\frac{1}{4}a^{2}+ab+b^{2}\right)$$

To find the opposite of \frac{1}{4}a^{2}+ab+b^{2}, find the opposite of each term.

$$\frac{1}{4}a^{2}-ab+b^{2}-\frac{1}{4}a^{2}-ab-b^{2}$$

Combine \frac{1}{4}a^{2} and -\frac{1}{4}a^{2} to get 0.

$$-ab+b^{2}-ab-b^{2}$$

Combine -ab and -ab to get -2ab.

$$-2ab+b^{2}-b^{2}$$

Combine b^{2} and -b^{2} to get 0.

$$-2ab$$

Expand

$$-2ab$$

Solution Steps

Use binomial theorem \left(p-q\right)^{2}=p^{2}-2pq+q^{2} to expand \left(\frac{1}{2}a-b\right)^{2}.

$$\frac{1}{4}a^{2}-ab+b^{2}-\left(\frac{1}{2}a+b\right)^{2}$$

Use binomial theorem \left(p+q\right)^{2}=p^{2}+2pq+q^{2} to expand \left(\frac{1}{2}a+b\right)^{2}.

$$\frac{1}{4}a^{2}-ab+b^{2}-\left(\frac{1}{4}a^{2}+ab+b^{2}\right)$$

To find the opposite of \frac{1}{4}a^{2}+ab+b^{2}, find the opposite of each term.

$$\frac{1}{4}a^{2}-ab+b^{2}-\frac{1}{4}a^{2}-ab-b^{2}$$

Combine \frac{1}{4}a^{2} and -\frac{1}{4}a^{2} to get 0.

$$-ab+b^{2}-ab-b^{2}$$

Combine -ab and -ab to get -2ab.

$$-2ab+b^{2}-b^{2}$$

Combine b^{2} and -b^{2} to get 0.

$$-2ab$$