Solve 25(a+b)^2-16(a-b)^2

$$25 ( a + b ) ^ { 2 } - 16 ( a - b ) ^ { 2 }$$

Evaluate

$$\left(a+9b\right)\left(9a+b\right)$$

Solution Steps

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

$$25\left(a^{2}+2ab+b^{2}\right)-16\left(a-b\right)^{2}$$

Use the distributive property to multiply 25 by a^{2}+2ab+b^{2}.

$$25a^{2}+50ab+25b^{2}-16\left(a-b\right)^{2}$$

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

$$25a^{2}+50ab+25b^{2}-16\left(a^{2}-2ab+b^{2}\right)$$

Use the distributive property to multiply -16 by a^{2}-2ab+b^{2}.

$$25a^{2}+50ab+25b^{2}-16a^{2}+32ab-16b^{2}$$

Combine 25a^{2} and -16a^{2} to get 9a^{2}.

$$9a^{2}+50ab+25b^{2}+32ab-16b^{2}$$

Combine 50ab and 32ab to get 82ab.

$$9a^{2}+82ab+25b^{2}-16b^{2}$$

Combine 25b^{2} and -16b^{2} to get 9b^{2}.

$$9a^{2}+82ab+9b^{2}$$

Expand

$$9a^{2}+82ab+9b^{2}$$

Solution Steps

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

$$25\left(a^{2}+2ab+b^{2}\right)-16\left(a-b\right)^{2}$$

Use the distributive property to multiply 25 by a^{2}+2ab+b^{2}.

$$25a^{2}+50ab+25b^{2}-16\left(a-b\right)^{2}$$

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

$$25a^{2}+50ab+25b^{2}-16\left(a^{2}-2ab+b^{2}\right)$$

Use the distributive property to multiply -16 by a^{2}-2ab+b^{2}.

$$25a^{2}+50ab+25b^{2}-16a^{2}+32ab-16b^{2}$$

Combine 25a^{2} and -16a^{2} to get 9a^{2}.

$$9a^{2}+50ab+25b^{2}+32ab-16b^{2}$$

Combine 50ab and 32ab to get 82ab.

$$9a^{2}+82ab+25b^{2}-16b^{2}$$

Combine 25b^{2} and -16b^{2} to get 9b^{2}.

$$9a^{2}+82ab+9b^{2}$$