Solve (1/6)^3+8(1/12)^3-1/6*1/12=

$$( \frac { 1 } { 6 } ) ^ { 3 } + 8 ( \frac { 1 } { 12 } ) ^ { 3 } - \frac { 1 } { 6 } \times \frac { 1 } { 12 } =$$

Evaluate

$$-\frac{1}{216}\approx -0.00462963$$

Solution Steps

Calculate \frac{1}{6} to the power of 3 and get \frac{1}{216}.

$$\frac{1}{216}+8\times \left(\frac{1}{12}\right)^{3}-\frac{1}{6}\times \frac{1}{12}$$

Calculate \frac{1}{12} to the power of 3 and get \frac{1}{1728}.

$$\frac{1}{216}+8\times \frac{1}{1728}-\frac{1}{6}\times \frac{1}{12}$$

Multiply 8 and \frac{1}{1728} to get \frac{8}{1728}.

$$\frac{1}{216}+\frac{8}{1728}-\frac{1}{6}\times \frac{1}{12}$$

Reduce the fraction \frac{8}{1728} to lowest terms by extracting and canceling out 8.

$$\frac{1}{216}+\frac{1}{216}-\frac{1}{6}\times \frac{1}{12}$$

Since \frac{1}{216} and \frac{1}{216} have the same denominator, add them by adding their numerators.

$$\frac{1+1}{216}-\frac{1}{6}\times \frac{1}{12}$$

Add 1 and 1 to get 2.

$$\frac{2}{216}-\frac{1}{6}\times \frac{1}{12}$$

Reduce the fraction \frac{2}{216} to lowest terms by extracting and canceling out 2.

$$\frac{1}{108}-\frac{1}{6}\times \frac{1}{12}$$

Multiply \frac{1}{6} times \frac{1}{12} by multiplying numerator times numerator and denominator times denominator.

$$\frac{1}{108}-\frac{1\times 1}{6\times 12}$$

Do the multiplications in the fraction \frac{1\times 1}{6\times 12}.

$$\frac{1}{108}-\frac{1}{72}$$

Least common multiple of 108 and 72 is 216. Convert \frac{1}{108} and \frac{1}{72} to fractions with denominator 216.

$$\frac{2}{216}-\frac{3}{216}$$

Since \frac{2}{216} and \frac{3}{216} have the same denominator, subtract them by subtracting their numerators.

$$\frac{2-3}{216}$$

Subtract 3 from 2 to get -1.

$$-\frac{1}{216}$$

Factor

$$-\frac{1}{216} = -0.004629629629629629$$