Solve (a-2b)^2-[(a-2b)(a+2b)-8b^2]=

$$( a - 2 b ) ^ { 2 } - [ ( a - 2 b ) ( a + 2 b ) - 8 b ^ { 2 } ] =$$

Evaluate

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

Solution Steps

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

$$a^{2}-4ab+4b^{2}-\left(\left(a-2b\right)\left(a+2b\right)-8b^{2}\right)$$

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

$$a^{2}-4ab+4b^{2}-\left(a^{2}-\left(2b\right)^{2}-8b^{2}\right)$$

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

$$a^{2}-4ab+4b^{2}-\left(a^{2}-2^{2}b^{2}-8b^{2}\right)$$

Calculate 2 to the power of 2 and get 4.

$$a^{2}-4ab+4b^{2}-\left(a^{2}-4b^{2}-8b^{2}\right)$$

Combine -4b^{2} and -8b^{2} to get -12b^{2}.

$$a^{2}-4ab+4b^{2}-\left(a^{2}-12b^{2}\right)$$

To find the opposite of a^{2}-12b^{2}, find the opposite of each term.

$$a^{2}-4ab+4b^{2}-a^{2}+12b^{2}$$

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

$$-4ab+4b^{2}+12b^{2}$$

Combine 4b^{2} and 12b^{2} to get 16b^{2}.

$$-4ab+16b^{2}$$

Expand

$$16b^{2}-4ab$$

Solution Steps

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

$$a^{2}-4ab+4b^{2}-\left(\left(a-2b\right)\left(a+2b\right)-8b^{2}\right)$$

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

$$a^{2}-4ab+4b^{2}-\left(a^{2}-\left(2b\right)^{2}-8b^{2}\right)$$

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

$$a^{2}-4ab+4b^{2}-\left(a^{2}-2^{2}b^{2}-8b^{2}\right)$$

Calculate 2 to the power of 2 and get 4.

$$a^{2}-4ab+4b^{2}-\left(a^{2}-4b^{2}-8b^{2}\right)$$

Combine -4b^{2} and -8b^{2} to get -12b^{2}.

$$a^{2}-4ab+4b^{2}-\left(a^{2}-12b^{2}\right)$$

To find the opposite of a^{2}-12b^{2}, find the opposite of each term.

$$a^{2}-4ab+4b^{2}-a^{2}+12b^{2}$$

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

$$-4ab+4b^{2}+12b^{2}$$

Combine 4b^{2} and 12b^{2} to get 16b^{2}.

$$-4ab+16b^{2}$$