Solve -1/4{left(x+1/2right)}^3-1/3

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

Evaluate

$$-\frac{\left(2x+1\right)^{3}}{32}-\frac{1}{3}$$

Solution Steps

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

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

Use the distributive property to multiply -\frac{1}{4} by x^{3}+\frac{3}{2}x^{2}+\frac{3}{4}x+\frac{1}{8}.

$$-\frac{1}{4}x^{3}-\frac{3}{8}x^{2}-\frac{3}{16}x-\frac{1}{32}-\frac{1}{3}$$

Subtract \frac{1}{3} from -\frac{1}{32} to get -\frac{35}{96}.

$$-\frac{1}{4}x^{3}-\frac{3}{8}x^{2}-\frac{3}{16}x-\frac{35}{96}$$

Expand

$$-\frac{x^{3}}{4}-\frac{3x^{2}}{8}-\frac{3x}{16}-\frac{35}{96}$$

Solution Steps

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

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

Use the distributive property to multiply -\frac{1}{4} by x^{3}+\frac{3}{2}x^{2}+\frac{3}{4}x+\frac{1}{8}.

$$-\frac{1}{4}x^{3}-\frac{3}{8}x^{2}-\frac{3}{16}x-\frac{1}{32}-\frac{1}{3}$$

Subtract \frac{1}{3} from -\frac{1}{32} to get -\frac{35}{96}.

$$-\frac{1}{4}x^{3}-\frac{3}{8}x^{2}-\frac{3}{16}x-\frac{35}{96}$$