Solve (a^2+1/2)^2-2(a^2+1/8)=

$$( a ^ { 2 } + \frac { 1 } { 2 } ) ^ { 2 } - 2 ( a ^ { 2 } + \frac { 1 } { 8 } ) =$$

Evaluate

$$a^{2}\left(a^{2}-1\right)$$

Solution Steps

Use binomial theorem \left(p+q\right)^{2}=p^{2}+2pq+q^{2} to expand \left(a^{2}+\frac{1}{2}\right)^{2}.

$$\left(a^{2}\right)^{2}+a^{2}+\frac{1}{4}-2\left(a^{2}+\frac{1}{8}\right)$$

To raise a power to another power, multiply the exponents. Multiply 2 and 2 to get 4.

$$a^{4}+a^{2}+\frac{1}{4}-2\left(a^{2}+\frac{1}{8}\right)$$

Use the distributive property to multiply -2 by a^{2}+\frac{1}{8}.

$$a^{4}+a^{2}+\frac{1}{4}-2a^{2}-\frac{1}{4}$$

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

$$a^{4}-a^{2}+\frac{1}{4}-\frac{1}{4}$$

Subtract \frac{1}{4} from \frac{1}{4} to get 0.

$$a^{4}-a^{2}$$

Expand

$$a^{4}-a^{2}$$

Solution Steps

Use binomial theorem \left(p+q\right)^{2}=p^{2}+2pq+q^{2} to expand \left(a^{2}+\frac{1}{2}\right)^{2}.

$$\left(a^{2}\right)^{2}+a^{2}+\frac{1}{4}-2\left(a^{2}+\frac{1}{8}\right)$$

To raise a power to another power, multiply the exponents. Multiply 2 and 2 to get 4.

$$a^{4}+a^{2}+\frac{1}{4}-2\left(a^{2}+\frac{1}{8}\right)$$

Use the distributive property to multiply -2 by a^{2}+\frac{1}{8}.

$$a^{4}+a^{2}+\frac{1}{4}-2a^{2}-\frac{1}{4}$$

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

$$a^{4}-a^{2}+\frac{1}{4}-\frac{1}{4}$$

Subtract \frac{1}{4} from \frac{1}{4} to get 0.

$$a^{4}-a^{2}$$