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

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

Evaluate

$$12ab$$

Solution Steps

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

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

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

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

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

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

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

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

Combine -3a^{2} and 3a^{2} to get 0.

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

Combine 6ab and 6ab to get 12ab.

$$12ab-3b^{2}+3b^{2}$$

Combine -3b^{2} and 3b^{2} to get 0.

$$12ab$$

Expand

$$12ab$$

Solution Steps

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

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

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

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

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

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

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

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

Combine -3a^{2} and 3a^{2} to get 0.

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

Combine 6ab and 6ab to get 12ab.

$$12ab-3b^{2}+3b^{2}$$

Combine -3b^{2} and 3b^{2} to get 0.

$$12ab$$