Solve [(a-b)^2-(a+b)^2]^2

$$[ ( a - b ) ^ { 2 } - ( a + b ) ^ { 2 } ] ^ { 2 }$$

Evaluate

$$16\left(ab\right)^{2}$$

Solution Steps

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

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

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

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

To find the opposite of a^{2}+2ab+b^{2}, find the opposite of each term.

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

Combine a^{2} and -a^{2} to get 0.

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

Combine -2ab and -2ab to get -4ab.

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

Combine b^{2} and -b^{2} to get 0.

$$\left(-4ab\right)^{2}$$

Expand \left(-4ab\right)^{2}.

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

Calculate -4 to the power of 2 and get 16.

$$16a^{2}b^{2}$$

Expand

$$16\left(ab\right)^{2}$$

Solution Steps

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

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

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

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

To find the opposite of a^{2}+2ab+b^{2}, find the opposite of each term.

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

Combine a^{2} and -a^{2} to get 0.

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

Combine -2ab and -2ab to get -4ab.

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

Combine b^{2} and -b^{2} to get 0.

$$\left(-4ab\right)^{2}$$

Expand \left(-4ab\right)^{2}.

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

Calculate -4 to the power of 2 and get 16.

$$16a^{2}b^{2}$$