Solve (-1/2a+b)(-1/2a-b)-(a+1/3b)(a-1/3b)

$$( - \frac { 1 } { 2 } a + b ) ( - \frac { 1 } { 2 } a - b ) - ( a + \frac { 1 } { 3 } b ) ( a - \frac { 1 } { 3 } b )$$

Evaluate

$$-\frac{3a^{2}}{4}-\frac{8b^{2}}{9}$$

Solution Steps

Consider \left(-\frac{1}{2}a+b\right)\left(-\frac{1}{2}a-b\right). Multiplication can be transformed into difference of squares using the rule: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}.

$$\left(-\frac{1}{2}a\right)^{2}-b^{2}-\left(a+\frac{1}{3}b\right)\left(a-\frac{1}{3}b\right)$$

Expand \left(-\frac{1}{2}a\right)^{2}.

$$\left(-\frac{1}{2}\right)^{2}a^{2}-b^{2}-\left(a+\frac{1}{3}b\right)\left(a-\frac{1}{3}b\right)$$

Calculate -\frac{1}{2} to the power of 2 and get \frac{1}{4}.

$$\frac{1}{4}a^{2}-b^{2}-\left(a+\frac{1}{3}b\right)\left(a-\frac{1}{3}b\right)$$

Consider \left(a+\frac{1}{3}b\right)\left(a-\frac{1}{3}b\right). Multiplication can be transformed into difference of squares using the rule: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}.

$$\frac{1}{4}a^{2}-b^{2}-\left(a^{2}-\left(\frac{1}{3}b\right)^{2}\right)$$

Expand \left(\frac{1}{3}b\right)^{2}.

$$\frac{1}{4}a^{2}-b^{2}-\left(a^{2}-\left(\frac{1}{3}\right)^{2}b^{2}\right)$$

Calculate \frac{1}{3} to the power of 2 and get \frac{1}{9}.

$$\frac{1}{4}a^{2}-b^{2}-\left(a^{2}-\frac{1}{9}b^{2}\right)$$

To find the opposite of a^{2}-\frac{1}{9}b^{2}, find the opposite of each term.

$$\frac{1}{4}a^{2}-b^{2}-a^{2}-\left(-\frac{1}{9}b^{2}\right)$$

The opposite of -\frac{1}{9}b^{2} is \frac{1}{9}b^{2}.

$$\frac{1}{4}a^{2}-b^{2}-a^{2}+\frac{1}{9}b^{2}$$

Combine \frac{1}{4}a^{2} and -a^{2} to get -\frac{3}{4}a^{2}.

$$-\frac{3}{4}a^{2}-b^{2}+\frac{1}{9}b^{2}$$

Combine -b^{2} and \frac{1}{9}b^{2} to get -\frac{8}{9}b^{2}.

$$-\frac{3}{4}a^{2}-\frac{8}{9}b^{2}$$

Expand

$$-\frac{3a^{2}}{4}-\frac{8b^{2}}{9}$$

Solution Steps

Consider \left(-\frac{1}{2}a+b\right)\left(-\frac{1}{2}a-b\right). Multiplication can be transformed into difference of squares using the rule: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}.

$$\left(-\frac{1}{2}a\right)^{2}-b^{2}-\left(a+\frac{1}{3}b\right)\left(a-\frac{1}{3}b\right)$$

Expand \left(-\frac{1}{2}a\right)^{2}.

$$\left(-\frac{1}{2}\right)^{2}a^{2}-b^{2}-\left(a+\frac{1}{3}b\right)\left(a-\frac{1}{3}b\right)$$

Calculate -\frac{1}{2} to the power of 2 and get \frac{1}{4}.

$$\frac{1}{4}a^{2}-b^{2}-\left(a+\frac{1}{3}b\right)\left(a-\frac{1}{3}b\right)$$

Consider \left(a+\frac{1}{3}b\right)\left(a-\frac{1}{3}b\right). Multiplication can be transformed into difference of squares using the rule: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}.

$$\frac{1}{4}a^{2}-b^{2}-\left(a^{2}-\left(\frac{1}{3}b\right)^{2}\right)$$

Expand \left(\frac{1}{3}b\right)^{2}.

$$\frac{1}{4}a^{2}-b^{2}-\left(a^{2}-\left(\frac{1}{3}\right)^{2}b^{2}\right)$$

Calculate \frac{1}{3} to the power of 2 and get \frac{1}{9}.

$$\frac{1}{4}a^{2}-b^{2}-\left(a^{2}-\frac{1}{9}b^{2}\right)$$

To find the opposite of a^{2}-\frac{1}{9}b^{2}, find the opposite of each term.

$$\frac{1}{4}a^{2}-b^{2}-a^{2}-\left(-\frac{1}{9}b^{2}\right)$$

The opposite of -\frac{1}{9}b^{2} is \frac{1}{9}b^{2}.

$$\frac{1}{4}a^{2}-b^{2}-a^{2}+\frac{1}{9}b^{2}$$

Combine \frac{1}{4}a^{2} and -a^{2} to get -\frac{3}{4}a^{2}.

$$-\frac{3}{4}a^{2}-b^{2}+\frac{1}{9}b^{2}$$

Combine -b^{2} and \frac{1}{9}b^{2} to get -\frac{8}{9}b^{2}.

$$-\frac{3}{4}a^{2}-\frac{8}{9}b^{2}$$