Solve (a+8)(a-8)+(a+4)^2=

$$( a + 8 ) ( a - 8 ) + ( a + 4 ) ^ { 2 } =$$

Evaluate

$$\left(a+4\right)^{2}+a^{2}-64$$

Solution Steps

Consider \left(a+8\right)\left(a-8\right). Multiplication can be transformed into difference of squares using the rule: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}. Square 8.

$$a^{2}-64+\left(a+4\right)^{2}$$

Use binomial theorem \left(p+q\right)^{2}=p^{2}+2pq+q^{2} to expand \left(a+4\right)^{2}.

$$a^{2}-64+a^{2}+8a+16$$

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

$$2a^{2}-64+8a+16$$

Add -64 and 16 to get -48.

$$2a^{2}-48+8a$$

Expand

$$2a^{2}+8a-48$$

Solution Steps

Consider \left(a+8\right)\left(a-8\right). Multiplication can be transformed into difference of squares using the rule: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}. Square 8.

$$a^{2}-64+\left(a+4\right)^{2}$$

Use binomial theorem \left(p+q\right)^{2}=p^{2}+2pq+q^{2} to expand \left(a+4\right)^{2}.

$$a^{2}-64+a^{2}+8a+16$$

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

$$2a^{2}-64+8a+16$$

Add -64 and 16 to get -48.

$$2a^{2}-48+8a$$