Solve 49(a-b)^2-16(a+b)^2

$$49 ( a - b ) ^ { 2 } - 16 ( a + b ) ^ { 2 }$$

Evaluate

$$\left(3a-11b\right)\left(11a-3b\right)$$

Solution Steps

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

$$49\left(a^{2}-2ab+b^{2}\right)-16\left(a+b\right)^{2}$$

Use the distributive property to multiply 49 by a^{2}-2ab+b^{2}.

$$49a^{2}-98ab+49b^{2}-16\left(a+b\right)^{2}$$

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

$$49a^{2}-98ab+49b^{2}-16\left(a^{2}+2ab+b^{2}\right)$$

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

$$49a^{2}-98ab+49b^{2}-16a^{2}-32ab-16b^{2}$$

Combine 49a^{2} and -16a^{2} to get 33a^{2}.

$$33a^{2}-98ab+49b^{2}-32ab-16b^{2}$$

Combine -98ab and -32ab to get -130ab.

$$33a^{2}-130ab+49b^{2}-16b^{2}$$

Combine 49b^{2} and -16b^{2} to get 33b^{2}.

$$33a^{2}-130ab+33b^{2}$$

Expand

$$33a^{2}-130ab+33b^{2}$$

Solution Steps

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

$$49\left(a^{2}-2ab+b^{2}\right)-16\left(a+b\right)^{2}$$

Use the distributive property to multiply 49 by a^{2}-2ab+b^{2}.

$$49a^{2}-98ab+49b^{2}-16\left(a+b\right)^{2}$$

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

$$49a^{2}-98ab+49b^{2}-16\left(a^{2}+2ab+b^{2}\right)$$

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

$$49a^{2}-98ab+49b^{2}-16a^{2}-32ab-16b^{2}$$

Combine 49a^{2} and -16a^{2} to get 33a^{2}.

$$33a^{2}-98ab+49b^{2}-32ab-16b^{2}$$

Combine -98ab and -32ab to get -130ab.

$$33a^{2}-130ab+49b^{2}-16b^{2}$$

Combine 49b^{2} and -16b^{2} to get 33b^{2}.

$$33a^{2}-130ab+33b^{2}$$