Solve (a-b)(a+b)-2(a-b)^2

$$( a - b ) ( a + b ) - 2 ( a - b ) ^ { 2 }$$

Evaluate

$$\left(a-b\right)\left(3b-a\right)$$

Solution Steps

Consider \left(a-b\right)\left(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}.

$$a^{2}-b^{2}-2\left(a-b\right)^{2}$$

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

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

Use the distributive property to multiply -2 by a^{2}-2ab+b^{2}.

$$a^{2}-b^{2}-2a^{2}+4ab-2b^{2}$$

Combine a^{2} and -2a^{2} to get -a^{2}.

$$-a^{2}-b^{2}+4ab-2b^{2}$$

Combine -b^{2} and -2b^{2} to get -3b^{2}.

$$-a^{2}-3b^{2}+4ab$$

Expand

$$-a^{2}+4ab-3b^{2}$$

Solution Steps

Consider \left(a-b\right)\left(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}.

$$a^{2}-b^{2}-2\left(a-b\right)^{2}$$

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

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

Use the distributive property to multiply -2 by a^{2}-2ab+b^{2}.

$$a^{2}-b^{2}-2a^{2}+4ab-2b^{2}$$

Combine a^{2} and -2a^{2} to get -a^{2}.

$$-a^{2}-b^{2}+4ab-2b^{2}$$

Combine -b^{2} and -2b^{2} to get -3b^{2}.

$$-a^{2}-3b^{2}+4ab$$