Solve (-1/2+1/3)*(-1/4*(1/4-frac{1}{8}

$$(- \frac{ 1 }{ 2 } + \frac{ 1 }{ 3 } ) \times (- \frac{ 1 }{ 4 } \times ( \frac{ 1 }{ 4 } - \frac{ 1 }{ 8 }$$

Evaluate

$$\frac{1}{192}\approx 0.005208333$$

Solution Steps

Least common multiple of 2 and 3 is 6. Convert -\frac{1}{2} and \frac{1}{3} to fractions with denominator 6.

$$\left(-\frac{3}{6}+\frac{2}{6}\right)\left(-\frac{1}{4}\right)\left(\frac{1}{4}-\frac{1}{8}\right)$$

Since -\frac{3}{6} and \frac{2}{6} have the same denominator, add them by adding their numerators.

$$\frac{-3+2}{6}\left(-\frac{1}{4}\right)\left(\frac{1}{4}-\frac{1}{8}\right)$$

Add -3 and 2 to get -1.

$$-\frac{1}{6}\left(-\frac{1}{4}\right)\left(\frac{1}{4}-\frac{1}{8}\right)$$

Multiply -\frac{1}{6} times -\frac{1}{4} by multiplying numerator times numerator and denominator times denominator.

$$\frac{-\left(-1\right)}{6\times 4}\left(\frac{1}{4}-\frac{1}{8}\right)$$

Do the multiplications in the fraction \frac{-\left(-1\right)}{6\times 4}.

$$\frac{1}{24}\left(\frac{1}{4}-\frac{1}{8}\right)$$

Least common multiple of 4 and 8 is 8. Convert \frac{1}{4} and \frac{1}{8} to fractions with denominator 8.

$$\frac{1}{24}\left(\frac{2}{8}-\frac{1}{8}\right)$$

Since \frac{2}{8} and \frac{1}{8} have the same denominator, subtract them by subtracting their numerators.

$$\frac{1}{24}\times \frac{2-1}{8}$$

Subtract 1 from 2 to get 1.

$$\frac{1}{24}\times \frac{1}{8}$$

Multiply \frac{1}{24} times \frac{1}{8} by multiplying numerator times numerator and denominator times denominator.

$$\frac{1\times 1}{24\times 8}$$

Do the multiplications in the fraction \frac{1\times 1}{24\times 8}.

$$\frac{1}{192}$$

Factor

$$\frac{1}{2 ^ {6} \cdot 3} = 0.005208333333333333$$