Solve 16a^2-9(a-b)^2

$$16 a ^ { 2 } - 9 ( a - b ) ^ { 2 }$$

Evaluate

$$\left(7a-3b\right)\left(a+3b\right)$$

Solution Steps

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

$$16a^{2}-9\left(a^{2}-2ab+b^{2}\right)$$

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

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

Combine 16a^{2} and -9a^{2} to get 7a^{2}.

$$7a^{2}+18ab-9b^{2}$$

Expand

$$7a^{2}+18ab-9b^{2}$$

Solution Steps

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

$$16a^{2}-9\left(a^{2}-2ab+b^{2}\right)$$

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

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

Combine 16a^{2} and -9a^{2} to get 7a^{2}.

$$7a^{2}+18ab-9b^{2}$$