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

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

Evaluate

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

Solution Steps

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

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

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

$$9a^{2}+18ab+9b^{2}-4\left(a-b\right)^{2}$$

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

$$9a^{2}+18ab+9b^{2}-4\left(a^{2}-2ab+b^{2}\right)$$

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

$$9a^{2}+18ab+9b^{2}-4a^{2}+8ab-4b^{2}$$

Combine 9a^{2} and -4a^{2} to get 5a^{2}.

$$5a^{2}+18ab+9b^{2}+8ab-4b^{2}$$

Combine 18ab and 8ab to get 26ab.

$$5a^{2}+26ab+9b^{2}-4b^{2}$$

Combine 9b^{2} and -4b^{2} to get 5b^{2}.

$$5a^{2}+26ab+5b^{2}$$

Expand

$$5a^{2}+26ab+5b^{2}$$

Solution Steps

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

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

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

$$9a^{2}+18ab+9b^{2}-4\left(a-b\right)^{2}$$

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

$$9a^{2}+18ab+9b^{2}-4\left(a^{2}-2ab+b^{2}\right)$$

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

$$9a^{2}+18ab+9b^{2}-4a^{2}+8ab-4b^{2}$$

Combine 9a^{2} and -4a^{2} to get 5a^{2}.

$$5a^{2}+18ab+9b^{2}+8ab-4b^{2}$$

Combine 18ab and 8ab to get 26ab.

$$5a^{2}+26ab+9b^{2}-4b^{2}$$

Combine 9b^{2} and -4b^{2} to get 5b^{2}.

$$5a^{2}+26ab+5b^{2}$$