Solve (4a-3b)^2+(b+2a)^2-10b^2-20a^2

$$( 4 a - 3 b ) ^ { 2 } + ( b + 2 a ) ^ { 2 } - 10 b ^ { 2 } - 20 a ^ { 2 }$$

Evaluate

$$-20ab$$

Solution Steps

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

$$16a^{2}-24ab+9b^{2}+\left(b+2a\right)^{2}-10b^{2}-20a^{2}$$

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

$$16a^{2}-24ab+9b^{2}+b^{2}+4ba+4a^{2}-10b^{2}-20a^{2}$$

Combine 9b^{2} and b^{2} to get 10b^{2}.

$$16a^{2}-24ab+10b^{2}+4ba+4a^{2}-10b^{2}-20a^{2}$$

Combine -24ab and 4ba to get -20ab.

$$16a^{2}-20ab+10b^{2}+4a^{2}-10b^{2}-20a^{2}$$

Combine 16a^{2} and 4a^{2} to get 20a^{2}.

$$20a^{2}-20ab+10b^{2}-10b^{2}-20a^{2}$$

Combine 10b^{2} and -10b^{2} to get 0.

$$20a^{2}-20ab-20a^{2}$$

Combine 20a^{2} and -20a^{2} to get 0.

$$-20ab$$

Expand

$$-20ab$$

Solution Steps

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

$$16a^{2}-24ab+9b^{2}+\left(b+2a\right)^{2}-10b^{2}-20a^{2}$$

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

$$16a^{2}-24ab+9b^{2}+b^{2}+4ba+4a^{2}-10b^{2}-20a^{2}$$

Combine 9b^{2} and b^{2} to get 10b^{2}.

$$16a^{2}-24ab+10b^{2}+4ba+4a^{2}-10b^{2}-20a^{2}$$

Combine -24ab and 4ba to get -20ab.

$$16a^{2}-20ab+10b^{2}+4a^{2}-10b^{2}-20a^{2}$$

Combine 16a^{2} and 4a^{2} to get 20a^{2}.

$$20a^{2}-20ab+10b^{2}-10b^{2}-20a^{2}$$

Combine 10b^{2} and -10b^{2} to get 0.

$$20a^{2}-20ab-20a^{2}$$

Combine 20a^{2} and -20a^{2} to get 0.

$$-20ab$$