Solve (v-1/2)^3-(1/2v-1)^3-7/8(v+1)(v-1)^2-(1/2v+frac{1}{8})^2-(v-frac{1}{8})(v+frac{1}{8})

$$( v - \frac { 1 } { 2 } ) ^ { 3 } - ( \frac { 1 } { 2 } v - 1 ) ^ { 3 } - \frac { 7 } { 8 } ( v + 1 ) ( v - 1 ) ^ { 2 } - ( \frac { 1 } { 2 } v + \frac { 1 } { 8 } ) ^ { 2 } - ( v - \frac { 1 } { 8 } ) ( v + \frac { 1 } { 8 } )$$

Evaluate

$$-\frac{9v^{2}}{8}$$

Solution Steps

Use binomial theorem \left(a-b\right)^{3}=a^{3}-3a^{2}b+3ab^{2}-b^{3} to expand \left(v-\frac{1}{2}\right)^{3}.

$$v^{3}-\frac{3}{2}v^{2}+\frac{3}{4}v-\frac{1}{8}-\left(\frac{1}{2}v-1\right)^{3}-\frac{7}{8}\left(v+1\right)\left(v-1\right)^{2}-\left(\frac{1}{2}v+\frac{1}{8}\right)^{2}-\left(v-\frac{1}{8}\right)\left(v+\frac{1}{8}\right)$$

Use binomial theorem \left(a-b\right)^{3}=a^{3}-3a^{2}b+3ab^{2}-b^{3} to expand \left(\frac{1}{2}v-1\right)^{3}.

$$v^{3}-\frac{3}{2}v^{2}+\frac{3}{4}v-\frac{1}{8}-\left(\frac{1}{8}v^{3}-\frac{3}{4}v^{2}+\frac{3}{2}v-1\right)-\frac{7}{8}\left(v+1\right)\left(v-1\right)^{2}-\left(\frac{1}{2}v+\frac{1}{8}\right)^{2}-\left(v-\frac{1}{8}\right)\left(v+\frac{1}{8}\right)$$

To find the opposite of \frac{1}{8}v^{3}-\frac{3}{4}v^{2}+\frac{3}{2}v-1, find the opposite of each term.

$$v^{3}-\frac{3}{2}v^{2}+\frac{3}{4}v-\frac{1}{8}-\frac{1}{8}v^{3}+\frac{3}{4}v^{2}-\frac{3}{2}v+1-\frac{7}{8}\left(v+1\right)\left(v-1\right)^{2}-\left(\frac{1}{2}v+\frac{1}{8}\right)^{2}-\left(v-\frac{1}{8}\right)\left(v+\frac{1}{8}\right)$$

Combine v^{3} and -\frac{1}{8}v^{3} to get \frac{7}{8}v^{3}.

$$\frac{7}{8}v^{3}-\frac{3}{2}v^{2}+\frac{3}{4}v-\frac{1}{8}+\frac{3}{4}v^{2}-\frac{3}{2}v+1-\frac{7}{8}\left(v+1\right)\left(v-1\right)^{2}-\left(\frac{1}{2}v+\frac{1}{8}\right)^{2}-\left(v-\frac{1}{8}\right)\left(v+\frac{1}{8}\right)$$

Combine -\frac{3}{2}v^{2} and \frac{3}{4}v^{2} to get -\frac{3}{4}v^{2}.

$$\frac{7}{8}v^{3}-\frac{3}{4}v^{2}+\frac{3}{4}v-\frac{1}{8}-\frac{3}{2}v+1-\frac{7}{8}\left(v+1\right)\left(v-1\right)^{2}-\left(\frac{1}{2}v+\frac{1}{8}\right)^{2}-\left(v-\frac{1}{8}\right)\left(v+\frac{1}{8}\right)$$

Combine \frac{3}{4}v and -\frac{3}{2}v to get -\frac{3}{4}v.

$$\frac{7}{8}v^{3}-\frac{3}{4}v^{2}-\frac{3}{4}v-\frac{1}{8}+1-\frac{7}{8}\left(v+1\right)\left(v-1\right)^{2}-\left(\frac{1}{2}v+\frac{1}{8}\right)^{2}-\left(v-\frac{1}{8}\right)\left(v+\frac{1}{8}\right)$$

Add -\frac{1}{8} and 1 to get \frac{7}{8}.

$$\frac{7}{8}v^{3}-\frac{3}{4}v^{2}-\frac{3}{4}v+\frac{7}{8}-\frac{7}{8}\left(v+1\right)\left(v-1\right)^{2}-\left(\frac{1}{2}v+\frac{1}{8}\right)^{2}-\left(v-\frac{1}{8}\right)\left(v+\frac{1}{8}\right)$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(v-1\right)^{2}.

$$\frac{7}{8}v^{3}-\frac{3}{4}v^{2}-\frac{3}{4}v+\frac{7}{8}-\frac{7}{8}\left(v+1\right)\left(v^{2}-2v+1\right)-\left(\frac{1}{2}v+\frac{1}{8}\right)^{2}-\left(v-\frac{1}{8}\right)\left(v+\frac{1}{8}\right)$$

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(\frac{1}{2}v+\frac{1}{8}\right)^{2}.

$$\frac{7}{8}v^{3}-\frac{3}{4}v^{2}-\frac{3}{4}v+\frac{7}{8}-\frac{7}{8}\left(v+1\right)\left(v^{2}-2v+1\right)-\left(\frac{1}{4}v^{2}+\frac{1}{8}v+\frac{1}{64}\right)-\left(v-\frac{1}{8}\right)\left(v+\frac{1}{8}\right)$$

To find the opposite of \frac{1}{4}v^{2}+\frac{1}{8}v+\frac{1}{64}, find the opposite of each term.

$$\frac{7}{8}v^{3}-\frac{3}{4}v^{2}-\frac{3}{4}v+\frac{7}{8}-\frac{7}{8}\left(v+1\right)\left(v^{2}-2v+1\right)-\frac{1}{4}v^{2}-\frac{1}{8}v-\frac{1}{64}-\left(v-\frac{1}{8}\right)\left(v+\frac{1}{8}\right)$$

Consider \left(v-\frac{1}{8}\right)\left(v+\frac{1}{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 \frac{1}{8}.

$$\frac{7}{8}v^{3}-\frac{3}{4}v^{2}-\frac{3}{4}v+\frac{7}{8}-\frac{7}{8}\left(v+1\right)\left(v^{2}-2v+1\right)-\frac{1}{4}v^{2}-\frac{1}{8}v-\frac{1}{64}-\left(v^{2}-\frac{1}{64}\right)$$

To find the opposite of v^{2}-\frac{1}{64}, find the opposite of each term.

$$\frac{7}{8}v^{3}-\frac{3}{4}v^{2}-\frac{3}{4}v+\frac{7}{8}-\frac{7}{8}\left(v+1\right)\left(v^{2}-2v+1\right)-\frac{1}{4}v^{2}-\frac{1}{8}v-\frac{1}{64}-v^{2}+\frac{1}{64}$$

Combine -\frac{1}{4}v^{2} and -v^{2} to get -\frac{5}{4}v^{2}.

$$\frac{7}{8}v^{3}-\frac{3}{4}v^{2}-\frac{3}{4}v+\frac{7}{8}-\frac{7}{8}\left(v+1\right)\left(v^{2}-2v+1\right)-\frac{5}{4}v^{2}-\frac{1}{8}v-\frac{1}{64}+\frac{1}{64}$$

Add -\frac{1}{64} and \frac{1}{64} to get 0.

$$\frac{7}{8}v^{3}-\frac{3}{4}v^{2}-\frac{3}{4}v+\frac{7}{8}-\frac{7}{8}\left(v+1\right)\left(v^{2}-2v+1\right)-\frac{5}{4}v^{2}-\frac{1}{8}v$$

Use the distributive property to multiply -\frac{7}{8} by v+1.

$$\frac{7}{8}v^{3}-\frac{3}{4}v^{2}-\frac{3}{4}v+\frac{7}{8}+\left(-\frac{7}{8}v-\frac{7}{8}\right)\left(v^{2}-2v+1\right)-\frac{5}{4}v^{2}-\frac{1}{8}v$$

Use the distributive property to multiply -\frac{7}{8}v-\frac{7}{8} by v^{2}-2v+1 and combine like terms.

$$\frac{7}{8}v^{3}-\frac{3}{4}v^{2}-\frac{3}{4}v+\frac{7}{8}-\frac{7}{8}v^{3}+\frac{7}{8}v^{2}+\frac{7}{8}v-\frac{7}{8}-\frac{5}{4}v^{2}-\frac{1}{8}v$$

Combine \frac{7}{8}v^{3} and -\frac{7}{8}v^{3} to get 0.

$$-\frac{3}{4}v^{2}-\frac{3}{4}v+\frac{7}{8}+\frac{7}{8}v^{2}+\frac{7}{8}v-\frac{7}{8}-\frac{5}{4}v^{2}-\frac{1}{8}v$$

Combine -\frac{3}{4}v^{2} and \frac{7}{8}v^{2} to get \frac{1}{8}v^{2}.

$$\frac{1}{8}v^{2}-\frac{3}{4}v+\frac{7}{8}+\frac{7}{8}v-\frac{7}{8}-\frac{5}{4}v^{2}-\frac{1}{8}v$$

Combine -\frac{3}{4}v and \frac{7}{8}v to get \frac{1}{8}v.

$$\frac{1}{8}v^{2}+\frac{1}{8}v+\frac{7}{8}-\frac{7}{8}-\frac{5}{4}v^{2}-\frac{1}{8}v$$

Subtract \frac{7}{8} from \frac{7}{8} to get 0.

$$\frac{1}{8}v^{2}+\frac{1}{8}v-\frac{5}{4}v^{2}-\frac{1}{8}v$$

Combine \frac{1}{8}v^{2} and -\frac{5}{4}v^{2} to get -\frac{9}{8}v^{2}.

$$-\frac{9}{8}v^{2}+\frac{1}{8}v-\frac{1}{8}v$$

Combine \frac{1}{8}v and -\frac{1}{8}v to get 0.

$$-\frac{9}{8}v^{2}$$

Expand

$$-\frac{9v^{2}}{8}$$

Solution Steps

Use binomial theorem \left(a-b\right)^{3}=a^{3}-3a^{2}b+3ab^{2}-b^{3} to expand \left(v-\frac{1}{2}\right)^{3}.

$$v^{3}-\frac{3}{2}v^{2}+\frac{3}{4}v-\frac{1}{8}-\left(\frac{1}{2}v-1\right)^{3}-\frac{7}{8}\left(v+1\right)\left(v-1\right)^{2}-\left(\frac{1}{2}v+\frac{1}{8}\right)^{2}-\left(v-\frac{1}{8}\right)\left(v+\frac{1}{8}\right)$$

Use binomial theorem \left(a-b\right)^{3}=a^{3}-3a^{2}b+3ab^{2}-b^{3} to expand \left(\frac{1}{2}v-1\right)^{3}.

$$v^{3}-\frac{3}{2}v^{2}+\frac{3}{4}v-\frac{1}{8}-\left(\frac{1}{8}v^{3}-\frac{3}{4}v^{2}+\frac{3}{2}v-1\right)-\frac{7}{8}\left(v+1\right)\left(v-1\right)^{2}-\left(\frac{1}{2}v+\frac{1}{8}\right)^{2}-\left(v-\frac{1}{8}\right)\left(v+\frac{1}{8}\right)$$

To find the opposite of \frac{1}{8}v^{3}-\frac{3}{4}v^{2}+\frac{3}{2}v-1, find the opposite of each term.

$$v^{3}-\frac{3}{2}v^{2}+\frac{3}{4}v-\frac{1}{8}-\frac{1}{8}v^{3}+\frac{3}{4}v^{2}-\frac{3}{2}v+1-\frac{7}{8}\left(v+1\right)\left(v-1\right)^{2}-\left(\frac{1}{2}v+\frac{1}{8}\right)^{2}-\left(v-\frac{1}{8}\right)\left(v+\frac{1}{8}\right)$$

Combine v^{3} and -\frac{1}{8}v^{3} to get \frac{7}{8}v^{3}.

$$\frac{7}{8}v^{3}-\frac{3}{2}v^{2}+\frac{3}{4}v-\frac{1}{8}+\frac{3}{4}v^{2}-\frac{3}{2}v+1-\frac{7}{8}\left(v+1\right)\left(v-1\right)^{2}-\left(\frac{1}{2}v+\frac{1}{8}\right)^{2}-\left(v-\frac{1}{8}\right)\left(v+\frac{1}{8}\right)$$

Combine -\frac{3}{2}v^{2} and \frac{3}{4}v^{2} to get -\frac{3}{4}v^{2}.

$$\frac{7}{8}v^{3}-\frac{3}{4}v^{2}+\frac{3}{4}v-\frac{1}{8}-\frac{3}{2}v+1-\frac{7}{8}\left(v+1\right)\left(v-1\right)^{2}-\left(\frac{1}{2}v+\frac{1}{8}\right)^{2}-\left(v-\frac{1}{8}\right)\left(v+\frac{1}{8}\right)$$

Combine \frac{3}{4}v and -\frac{3}{2}v to get -\frac{3}{4}v.

$$\frac{7}{8}v^{3}-\frac{3}{4}v^{2}-\frac{3}{4}v-\frac{1}{8}+1-\frac{7}{8}\left(v+1\right)\left(v-1\right)^{2}-\left(\frac{1}{2}v+\frac{1}{8}\right)^{2}-\left(v-\frac{1}{8}\right)\left(v+\frac{1}{8}\right)$$

Add -\frac{1}{8} and 1 to get \frac{7}{8}.

$$\frac{7}{8}v^{3}-\frac{3}{4}v^{2}-\frac{3}{4}v+\frac{7}{8}-\frac{7}{8}\left(v+1\right)\left(v-1\right)^{2}-\left(\frac{1}{2}v+\frac{1}{8}\right)^{2}-\left(v-\frac{1}{8}\right)\left(v+\frac{1}{8}\right)$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(v-1\right)^{2}.

$$\frac{7}{8}v^{3}-\frac{3}{4}v^{2}-\frac{3}{4}v+\frac{7}{8}-\frac{7}{8}\left(v+1\right)\left(v^{2}-2v+1\right)-\left(\frac{1}{2}v+\frac{1}{8}\right)^{2}-\left(v-\frac{1}{8}\right)\left(v+\frac{1}{8}\right)$$

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(\frac{1}{2}v+\frac{1}{8}\right)^{2}.

$$\frac{7}{8}v^{3}-\frac{3}{4}v^{2}-\frac{3}{4}v+\frac{7}{8}-\frac{7}{8}\left(v+1\right)\left(v^{2}-2v+1\right)-\left(\frac{1}{4}v^{2}+\frac{1}{8}v+\frac{1}{64}\right)-\left(v-\frac{1}{8}\right)\left(v+\frac{1}{8}\right)$$

To find the opposite of \frac{1}{4}v^{2}+\frac{1}{8}v+\frac{1}{64}, find the opposite of each term.

$$\frac{7}{8}v^{3}-\frac{3}{4}v^{2}-\frac{3}{4}v+\frac{7}{8}-\frac{7}{8}\left(v+1\right)\left(v^{2}-2v+1\right)-\frac{1}{4}v^{2}-\frac{1}{8}v-\frac{1}{64}-\left(v-\frac{1}{8}\right)\left(v+\frac{1}{8}\right)$$

Consider \left(v-\frac{1}{8}\right)\left(v+\frac{1}{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 \frac{1}{8}.

$$\frac{7}{8}v^{3}-\frac{3}{4}v^{2}-\frac{3}{4}v+\frac{7}{8}-\frac{7}{8}\left(v+1\right)\left(v^{2}-2v+1\right)-\frac{1}{4}v^{2}-\frac{1}{8}v-\frac{1}{64}-\left(v^{2}-\frac{1}{64}\right)$$

To find the opposite of v^{2}-\frac{1}{64}, find the opposite of each term.

$$\frac{7}{8}v^{3}-\frac{3}{4}v^{2}-\frac{3}{4}v+\frac{7}{8}-\frac{7}{8}\left(v+1\right)\left(v^{2}-2v+1\right)-\frac{1}{4}v^{2}-\frac{1}{8}v-\frac{1}{64}-v^{2}+\frac{1}{64}$$

Combine -\frac{1}{4}v^{2} and -v^{2} to get -\frac{5}{4}v^{2}.

$$\frac{7}{8}v^{3}-\frac{3}{4}v^{2}-\frac{3}{4}v+\frac{7}{8}-\frac{7}{8}\left(v+1\right)\left(v^{2}-2v+1\right)-\frac{5}{4}v^{2}-\frac{1}{8}v-\frac{1}{64}+\frac{1}{64}$$

Add -\frac{1}{64} and \frac{1}{64} to get 0.

$$\frac{7}{8}v^{3}-\frac{3}{4}v^{2}-\frac{3}{4}v+\frac{7}{8}-\frac{7}{8}\left(v+1\right)\left(v^{2}-2v+1\right)-\frac{5}{4}v^{2}-\frac{1}{8}v$$

Use the distributive property to multiply -\frac{7}{8} by v+1.

$$\frac{7}{8}v^{3}-\frac{3}{4}v^{2}-\frac{3}{4}v+\frac{7}{8}+\left(-\frac{7}{8}v-\frac{7}{8}\right)\left(v^{2}-2v+1\right)-\frac{5}{4}v^{2}-\frac{1}{8}v$$

Use the distributive property to multiply -\frac{7}{8}v-\frac{7}{8} by v^{2}-2v+1 and combine like terms.

$$\frac{7}{8}v^{3}-\frac{3}{4}v^{2}-\frac{3}{4}v+\frac{7}{8}-\frac{7}{8}v^{3}+\frac{7}{8}v^{2}+\frac{7}{8}v-\frac{7}{8}-\frac{5}{4}v^{2}-\frac{1}{8}v$$

Combine \frac{7}{8}v^{3} and -\frac{7}{8}v^{3} to get 0.

$$-\frac{3}{4}v^{2}-\frac{3}{4}v+\frac{7}{8}+\frac{7}{8}v^{2}+\frac{7}{8}v-\frac{7}{8}-\frac{5}{4}v^{2}-\frac{1}{8}v$$

Combine -\frac{3}{4}v^{2} and \frac{7}{8}v^{2} to get \frac{1}{8}v^{2}.

$$\frac{1}{8}v^{2}-\frac{3}{4}v+\frac{7}{8}+\frac{7}{8}v-\frac{7}{8}-\frac{5}{4}v^{2}-\frac{1}{8}v$$

Combine -\frac{3}{4}v and \frac{7}{8}v to get \frac{1}{8}v.

$$\frac{1}{8}v^{2}+\frac{1}{8}v+\frac{7}{8}-\frac{7}{8}-\frac{5}{4}v^{2}-\frac{1}{8}v$$

Subtract \frac{7}{8} from \frac{7}{8} to get 0.

$$\frac{1}{8}v^{2}+\frac{1}{8}v-\frac{5}{4}v^{2}-\frac{1}{8}v$$

Combine \frac{1}{8}v^{2} and -\frac{5}{4}v^{2} to get -\frac{9}{8}v^{2}.

$$-\frac{9}{8}v^{2}+\frac{1}{8}v-\frac{1}{8}v$$

Combine \frac{1}{8}v and -\frac{1}{8}v to get 0.

$$-\frac{9}{8}v^{2}$$