Solve 3*(1/3)^3-3(1/3)^2+1/3+1

$$3 \cdot ( \frac { 1 } { 3 } ) ^ { 3 } - 3 ( \frac { 1 } { 3 } ) ^ { 2 } + \frac { 1 } { 3 } + 1$$

Evaluate

$$\frac{10}{9}\approx 1.111111111$$

Solution Steps

Calculate \frac{1}{3} to the power of 3 and get \frac{1}{27}.

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

Multiply 3 and \frac{1}{27} to get \frac{3}{27}.

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

Reduce the fraction \frac{3}{27} to lowest terms by extracting and canceling out 3.

$$\frac{1}{9}-3\times \left(\frac{1}{3}\right)^{2}+\frac{1}{3}+1$$

Calculate \frac{1}{3} to the power of 2 and get \frac{1}{9}.

$$\frac{1}{9}-3\times \frac{1}{9}+\frac{1}{3}+1$$

Multiply 3 and \frac{1}{9} to get \frac{3}{9}.

$$\frac{1}{9}-\frac{3}{9}+\frac{1}{3}+1$$

Since \frac{1}{9} and \frac{3}{9} have the same denominator, subtract them by subtracting their numerators.

$$\frac{1-3}{9}+\frac{1}{3}+1$$

Subtract 3 from 1 to get -2.

$$-\frac{2}{9}+\frac{1}{3}+1$$

Least common multiple of 9 and 3 is 9. Convert -\frac{2}{9} and \frac{1}{3} to fractions with denominator 9.

$$-\frac{2}{9}+\frac{3}{9}+1$$

Since -\frac{2}{9} and \frac{3}{9} have the same denominator, add them by adding their numerators.

$$\frac{-2+3}{9}+1$$

Add -2 and 3 to get 1.

$$\frac{1}{9}+1$$

Convert 1 to fraction \frac{9}{9}.

$$\frac{1}{9}+\frac{9}{9}$$

Since \frac{1}{9} and \frac{9}{9} have the same denominator, add them by adding their numerators.

$$\frac{1+9}{9}$$

Add 1 and 9 to get 10.

$$\frac{10}{9}$$

Factor

$$\frac{2 \cdot 5}{3 ^ {2}} = 1\frac{1}{9} = 1.1111111111111112$$