Solve +1/8{left(x-1/3right)}^3-1/27

$$+ \frac{ 1 }{ 8 } { \left(x- \frac{ 1 }{ 3 } \right) }^{ 3 } - \frac{ 1 }{ 27 }$$

Evaluate

$$\frac{\left(x-1\right)\left(3x^{2}+1\right)}{24}$$

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}{3}\right)^{3}.

$$\frac{1}{8}\left(x^{3}-x^{2}+\frac{1}{3}x-\frac{1}{27}\right)-\frac{1}{27}$$

Use the distributive property to multiply \frac{1}{8} by x^{3}-x^{2}+\frac{1}{3}x-\frac{1}{27}.

$$\frac{1}{8}x^{3}-\frac{1}{8}x^{2}+\frac{1}{24}x-\frac{1}{216}-\frac{1}{27}$$

Subtract \frac{1}{27} from -\frac{1}{216} to get -\frac{1}{24}.

$$\frac{1}{8}x^{3}-\frac{1}{8}x^{2}+\frac{1}{24}x-\frac{1}{24}$$

Expand

$$\frac{x^{3}}{8}-\frac{x^{2}}{8}+\frac{x}{24}-\frac{1}{24}$$

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}{3}\right)^{3}.

$$\frac{1}{8}\left(x^{3}-x^{2}+\frac{1}{3}x-\frac{1}{27}\right)-\frac{1}{27}$$

Use the distributive property to multiply \frac{1}{8} by x^{3}-x^{2}+\frac{1}{3}x-\frac{1}{27}.

$$\frac{1}{8}x^{3}-\frac{1}{8}x^{2}+\frac{1}{24}x-\frac{1}{216}-\frac{1}{27}$$

Subtract \frac{1}{27} from -\frac{1}{216} to get -\frac{1}{24}.

$$\frac{1}{8}x^{3}-\frac{1}{8}x^{2}+\frac{1}{24}x-\frac{1}{24}$$