Solve 9(a+b)^2-36(a-b)^2

$$9 ( a + b ) ^ { 2 } - 36 ( a - b ) ^ { 2 }$$

Evaluate

$$9\left(3a-b\right)\left(3b-a\right)$$

Solution Steps

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

$$9\left(a^{2}+2ab+b^{2}\right)-36\left(a-b\right)^{2}$$

Use the distributive property to multiply 9 by a^{2}+2ab+b^{2}.

$$9a^{2}+18ab+9b^{2}-36\left(a-b\right)^{2}$$

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

$$9a^{2}+18ab+9b^{2}-36\left(a^{2}-2ab+b^{2}\right)$$

Use the distributive property to multiply -36 by a^{2}-2ab+b^{2}.

$$9a^{2}+18ab+9b^{2}-36a^{2}+72ab-36b^{2}$$

Combine 9a^{2} and -36a^{2} to get -27a^{2}.

$$-27a^{2}+18ab+9b^{2}+72ab-36b^{2}$$

Combine 18ab and 72ab to get 90ab.

$$-27a^{2}+90ab+9b^{2}-36b^{2}$$

Combine 9b^{2} and -36b^{2} to get -27b^{2}.

$$-27a^{2}+90ab-27b^{2}$$

Expand

$$-27a^{2}+90ab-27b^{2}$$

Solution Steps

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

$$9\left(a^{2}+2ab+b^{2}\right)-36\left(a-b\right)^{2}$$

Use the distributive property to multiply 9 by a^{2}+2ab+b^{2}.

$$9a^{2}+18ab+9b^{2}-36\left(a-b\right)^{2}$$

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

$$9a^{2}+18ab+9b^{2}-36\left(a^{2}-2ab+b^{2}\right)$$

Use the distributive property to multiply -36 by a^{2}-2ab+b^{2}.

$$9a^{2}+18ab+9b^{2}-36a^{2}+72ab-36b^{2}$$

Combine 9a^{2} and -36a^{2} to get -27a^{2}.

$$-27a^{2}+18ab+9b^{2}+72ab-36b^{2}$$

Combine 18ab and 72ab to get 90ab.

$$-27a^{2}+90ab+9b^{2}-36b^{2}$$

Combine 9b^{2} and -36b^{2} to get -27b^{2}.

$$-27a^{2}+90ab-27b^{2}$$