Solve [(1/3+6)-(-8/3+12)]1/3

$$[ ( \frac { 1 } { 3 } + 6 ) - ( \frac { - 8 } { 3 } + 12 ) ] \frac { 1 } { 3 }$$

Evaluate

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Solution Steps

Convert 6 to fraction \frac{18}{3}.

$$\left(\frac{1}{3}+\frac{18}{3}-\left(\frac{-8}{3}+12\right)\right)\times \frac{1}{3}$$

Since \frac{1}{3} and \frac{18}{3} have the same denominator, add them by adding their numerators.

$$\left(\frac{1+18}{3}-\left(\frac{-8}{3}+12\right)\right)\times \frac{1}{3}$$

Add 1 and 18 to get 19.

$$\left(\frac{19}{3}-\left(\frac{-8}{3}+12\right)\right)\times \frac{1}{3}$$

Fraction \frac{-8}{3} can be rewritten as -\frac{8}{3} by extracting the negative sign.

$$\left(\frac{19}{3}-\left(-\frac{8}{3}+12\right)\right)\times \frac{1}{3}$$

Convert 12 to fraction \frac{36}{3}.

$$\left(\frac{19}{3}-\left(-\frac{8}{3}+\frac{36}{3}\right)\right)\times \frac{1}{3}$$

Since -\frac{8}{3} and \frac{36}{3} have the same denominator, add them by adding their numerators.

$$\left(\frac{19}{3}-\frac{-8+36}{3}\right)\times \frac{1}{3}$$

Add -8 and 36 to get 28.

$$\left(\frac{19}{3}-\frac{28}{3}\right)\times \frac{1}{3}$$

Since \frac{19}{3} and \frac{28}{3} have the same denominator, subtract them by subtracting their numerators.

$$\frac{19-28}{3}\times \frac{1}{3}$$

Subtract 28 from 19 to get -9.

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

Divide -9 by 3 to get -3.

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

Multiply -3 times \frac{1}{3}.

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Factor

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