Solve (3a-4b)^2-(3a-b)^2

$$( 3 a - 4 b ) ^ { 2 } - ( 3 a - b ) ^ { 2 }$$

Evaluate

$$-3b\left(6a-5b\right)$$

Solution Steps

Use binomial theorem \left(p-q\right)^{2}=p^{2}-2pq+q^{2} to expand \left(3a-4b\right)^{2}.

$$9a^{2}-24ab+16b^{2}-\left(3a-b\right)^{2}$$

Use binomial theorem \left(p-q\right)^{2}=p^{2}-2pq+q^{2} to expand \left(3a-b\right)^{2}.

$$9a^{2}-24ab+16b^{2}-\left(9a^{2}-6ab+b^{2}\right)$$

To find the opposite of 9a^{2}-6ab+b^{2}, find the opposite of each term.

$$9a^{2}-24ab+16b^{2}-9a^{2}+6ab-b^{2}$$

Combine 9a^{2} and -9a^{2} to get 0.

$$-24ab+16b^{2}+6ab-b^{2}$$

Combine -24ab and 6ab to get -18ab.

$$-18ab+16b^{2}-b^{2}$$

Combine 16b^{2} and -b^{2} to get 15b^{2}.

$$-18ab+15b^{2}$$

Expand

$$15b^{2}-18ab$$

Solution Steps

Use binomial theorem \left(p-q\right)^{2}=p^{2}-2pq+q^{2} to expand \left(3a-4b\right)^{2}.

$$9a^{2}-24ab+16b^{2}-\left(3a-b\right)^{2}$$

Use binomial theorem \left(p-q\right)^{2}=p^{2}-2pq+q^{2} to expand \left(3a-b\right)^{2}.

$$9a^{2}-24ab+16b^{2}-\left(9a^{2}-6ab+b^{2}\right)$$

To find the opposite of 9a^{2}-6ab+b^{2}, find the opposite of each term.

$$9a^{2}-24ab+16b^{2}-9a^{2}+6ab-b^{2}$$

Combine 9a^{2} and -9a^{2} to get 0.

$$-24ab+16b^{2}+6ab-b^{2}$$

Combine -24ab and 6ab to get -18ab.

$$-18ab+16b^{2}-b^{2}$$

Combine 16b^{2} and -b^{2} to get 15b^{2}.

$$-18ab+15b^{2}$$