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

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

Evaluate

$$b\left(3b-4a\right)$$

Solution Steps

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

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

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}.

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

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

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

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

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

Combine 4b^{2} and b^{2} to get 5b^{2}.

$$-4ab+5b^{2}-2b^{2}$$

Combine 5b^{2} and -2b^{2} to get 3b^{2}.

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

Expand

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

Solution Steps

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

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

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}.

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

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

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

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

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

Combine 4b^{2} and b^{2} to get 5b^{2}.

$$-4ab+5b^{2}-2b^{2}$$

Combine 5b^{2} and -2b^{2} to get 3b^{2}.

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