Solve 9(2a+3b)^2-4(3a-2b)^2

$$9 ( 2 a + 3 b ) ^ { 2 } - 4 ( 3 a - 2 b ) ^ { 2 }$$

Evaluate

$$13b\left(12a+5b\right)$$

Solution Steps

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

$$9\left(4a^{2}+12ab+9b^{2}\right)-4\left(3a-2b\right)^{2}$$

Use the distributive property to multiply 9 by 4a^{2}+12ab+9b^{2}.

$$36a^{2}+108ab+81b^{2}-4\left(3a-2b\right)^{2}$$

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

$$36a^{2}+108ab+81b^{2}-4\left(9a^{2}-12ab+4b^{2}\right)$$

Use the distributive property to multiply -4 by 9a^{2}-12ab+4b^{2}.

$$36a^{2}+108ab+81b^{2}-36a^{2}+48ab-16b^{2}$$

Combine 36a^{2} and -36a^{2} to get 0.

$$108ab+81b^{2}+48ab-16b^{2}$$

Combine 108ab and 48ab to get 156ab.

$$156ab+81b^{2}-16b^{2}$$

Combine 81b^{2} and -16b^{2} to get 65b^{2}.

$$156ab+65b^{2}$$

Expand

$$156ab+65b^{2}$$

Solution Steps

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

$$9\left(4a^{2}+12ab+9b^{2}\right)-4\left(3a-2b\right)^{2}$$

Use the distributive property to multiply 9 by 4a^{2}+12ab+9b^{2}.

$$36a^{2}+108ab+81b^{2}-4\left(3a-2b\right)^{2}$$

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

$$36a^{2}+108ab+81b^{2}-4\left(9a^{2}-12ab+4b^{2}\right)$$

Use the distributive property to multiply -4 by 9a^{2}-12ab+4b^{2}.

$$36a^{2}+108ab+81b^{2}-36a^{2}+48ab-16b^{2}$$

Combine 36a^{2} and -36a^{2} to get 0.

$$108ab+81b^{2}+48ab-16b^{2}$$

Combine 108ab and 48ab to get 156ab.

$$156ab+81b^{2}-16b^{2}$$

Combine 81b^{2} and -16b^{2} to get 65b^{2}.

$$156ab+65b^{2}$$