Solve (a-5)^2-(a-8)^2

$$( a - 5 ) ^ { 2 } - ( a - 8 ) ^ { 2 }$$

Evaluate

$$6a-39$$

Solution Steps

Use binomial theorem \left(p-q\right)^{2}=p^{2}-2pq+q^{2} to expand \left(a-5\right)^{2}.

$$a^{2}-10a+25-\left(a-8\right)^{2}$$

Use binomial theorem \left(p-q\right)^{2}=p^{2}-2pq+q^{2} to expand \left(a-8\right)^{2}.

$$a^{2}-10a+25-\left(a^{2}-16a+64\right)$$

To find the opposite of a^{2}-16a+64, find the opposite of each term.

$$a^{2}-10a+25-a^{2}+16a-64$$

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

$$-10a+25+16a-64$$

Combine -10a and 16a to get 6a.

$$6a+25-64$$

Subtract 64 from 25 to get -39.

$$6a-39$$

Expand

$$6a-39$$

Solution Steps

Use binomial theorem \left(p-q\right)^{2}=p^{2}-2pq+q^{2} to expand \left(a-5\right)^{2}.

$$a^{2}-10a+25-\left(a-8\right)^{2}$$

Use binomial theorem \left(p-q\right)^{2}=p^{2}-2pq+q^{2} to expand \left(a-8\right)^{2}.

$$a^{2}-10a+25-\left(a^{2}-16a+64\right)$$

To find the opposite of a^{2}-16a+64, find the opposite of each term.

$$a^{2}-10a+25-a^{2}+16a-64$$

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

$$-10a+25+16a-64$$

Combine -10a and 16a to get 6a.

$$6a+25-64$$

Subtract 64 from 25 to get -39.

$$6a-39$$