Solve 25(a-b)^2-9(a+b)^2

$$25 ( a - b ) ^ { 2 } - 9 ( a + b ) ^ { 2 }$$

Evaluate

$$4\left(a-4b\right)\left(4a-b\right)$$

Solution Steps

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

$$25\left(a^{2}-2ab+b^{2}\right)-9\left(a+b\right)^{2}$$

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

$$25a^{2}-50ab+25b^{2}-9\left(a+b\right)^{2}$$

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

$$25a^{2}-50ab+25b^{2}-9\left(a^{2}+2ab+b^{2}\right)$$

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

$$25a^{2}-50ab+25b^{2}-9a^{2}-18ab-9b^{2}$$

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

$$16a^{2}-50ab+25b^{2}-18ab-9b^{2}$$

Combine -50ab and -18ab to get -68ab.

$$16a^{2}-68ab+25b^{2}-9b^{2}$$

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

$$16a^{2}-68ab+16b^{2}$$

Expand

$$16a^{2}-68ab+16b^{2}$$

Solution Steps

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

$$25\left(a^{2}-2ab+b^{2}\right)-9\left(a+b\right)^{2}$$

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

$$25a^{2}-50ab+25b^{2}-9\left(a+b\right)^{2}$$

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

$$25a^{2}-50ab+25b^{2}-9\left(a^{2}+2ab+b^{2}\right)$$

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

$$25a^{2}-50ab+25b^{2}-9a^{2}-18ab-9b^{2}$$

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

$$16a^{2}-50ab+25b^{2}-18ab-9b^{2}$$

Combine -50ab and -18ab to get -68ab.

$$16a^{2}-68ab+25b^{2}-9b^{2}$$

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

$$16a^{2}-68ab+16b^{2}$$