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

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

Evaluate

$$6b^{2}-2a^{2}$$

Solution Steps

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

$$a^{2}+6ab+9b^{2}-3\left(a+b\right)^{2}$$

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

$$a^{2}+6ab+9b^{2}-3\left(a^{2}+2ab+b^{2}\right)$$

Use the distributive property to multiply -3 by a^{2}+2ab+b^{2}.

$$a^{2}+6ab+9b^{2}-3a^{2}-6ab-3b^{2}$$

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

$$-2a^{2}+6ab+9b^{2}-6ab-3b^{2}$$

Combine 6ab and -6ab to get 0.

$$-2a^{2}+9b^{2}-3b^{2}$$

Combine 9b^{2} and -3b^{2} to get 6b^{2}.

$$-2a^{2}+6b^{2}$$

Expand

$$6b^{2}-2a^{2}$$

Solution Steps

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

$$a^{2}+6ab+9b^{2}-3\left(a+b\right)^{2}$$

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

$$a^{2}+6ab+9b^{2}-3\left(a^{2}+2ab+b^{2}\right)$$

Use the distributive property to multiply -3 by a^{2}+2ab+b^{2}.

$$a^{2}+6ab+9b^{2}-3a^{2}-6ab-3b^{2}$$

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

$$-2a^{2}+6ab+9b^{2}-6ab-3b^{2}$$

Combine 6ab and -6ab to get 0.

$$-2a^{2}+9b^{2}-3b^{2}$$

Combine 9b^{2} and -3b^{2} to get 6b^{2}.

$$-2a^{2}+6b^{2}$$