Solve 6[72+1/12(2^3-2)+1/12(3^3-3)+1/12(2^3-2)]/10-(10^2-1)

$$\frac { 6 [ 72 + \frac { 1 } { 12 } ( 2 ^ { 3 } - 2 ) + \frac { 1 } { 12 } ( 3 ^ { 3 } - 3 ) + \frac { 1 } { 12 } ( 2 ^ { 3 } - 2 ) ] } { 10 - ( 10 ^ { 2 } - 1 ) }$$

Evaluate

$$-\frac{450}{89}\approx -5.056179775$$

Solution Steps

Calculate 2 to the power of 3 and get 8.

$$\frac{6\left(72+\frac{1}{12}\left(8-2\right)+\frac{1}{12}\left(3^{3}-3\right)+\frac{1}{12}\left(2^{3}-2\right)\right)}{10-\left(10^{2}-1\right)}$$

Subtract 2 from 8 to get 6.

$$\frac{6\left(72+\frac{1}{12}\times 6+\frac{1}{12}\left(3^{3}-3\right)+\frac{1}{12}\left(2^{3}-2\right)\right)}{10-\left(10^{2}-1\right)}$$

Multiply \frac{1}{12} and 6 to get \frac{6}{12}.

$$\frac{6\left(72+\frac{6}{12}+\frac{1}{12}\left(3^{3}-3\right)+\frac{1}{12}\left(2^{3}-2\right)\right)}{10-\left(10^{2}-1\right)}$$

Reduce the fraction \frac{6}{12} to lowest terms by extracting and canceling out 6.

$$\frac{6\left(72+\frac{1}{2}+\frac{1}{12}\left(3^{3}-3\right)+\frac{1}{12}\left(2^{3}-2\right)\right)}{10-\left(10^{2}-1\right)}$$

Convert 72 to fraction \frac{144}{2}.

$$\frac{6\left(\frac{144}{2}+\frac{1}{2}+\frac{1}{12}\left(3^{3}-3\right)+\frac{1}{12}\left(2^{3}-2\right)\right)}{10-\left(10^{2}-1\right)}$$

Since \frac{144}{2} and \frac{1}{2} have the same denominator, add them by adding their numerators.

$$\frac{6\left(\frac{144+1}{2}+\frac{1}{12}\left(3^{3}-3\right)+\frac{1}{12}\left(2^{3}-2\right)\right)}{10-\left(10^{2}-1\right)}$$

Add 144 and 1 to get 145.

$$\frac{6\left(\frac{145}{2}+\frac{1}{12}\left(3^{3}-3\right)+\frac{1}{12}\left(2^{3}-2\right)\right)}{10-\left(10^{2}-1\right)}$$

Calculate 3 to the power of 3 and get 27.

$$\frac{6\left(\frac{145}{2}+\frac{1}{12}\left(27-3\right)+\frac{1}{12}\left(2^{3}-2\right)\right)}{10-\left(10^{2}-1\right)}$$

Subtract 3 from 27 to get 24.

$$\frac{6\left(\frac{145}{2}+\frac{1}{12}\times 24+\frac{1}{12}\left(2^{3}-2\right)\right)}{10-\left(10^{2}-1\right)}$$

Multiply \frac{1}{12} and 24 to get \frac{24}{12}.

$$\frac{6\left(\frac{145}{2}+\frac{24}{12}+\frac{1}{12}\left(2^{3}-2\right)\right)}{10-\left(10^{2}-1\right)}$$

Divide 24 by 12 to get 2.

$$\frac{6\left(\frac{145}{2}+2+\frac{1}{12}\left(2^{3}-2\right)\right)}{10-\left(10^{2}-1\right)}$$

Convert 2 to fraction \frac{4}{2}.

$$\frac{6\left(\frac{145}{2}+\frac{4}{2}+\frac{1}{12}\left(2^{3}-2\right)\right)}{10-\left(10^{2}-1\right)}$$

Since \frac{145}{2} and \frac{4}{2} have the same denominator, add them by adding their numerators.

$$\frac{6\left(\frac{145+4}{2}+\frac{1}{12}\left(2^{3}-2\right)\right)}{10-\left(10^{2}-1\right)}$$

Add 145 and 4 to get 149.

$$\frac{6\left(\frac{149}{2}+\frac{1}{12}\left(2^{3}-2\right)\right)}{10-\left(10^{2}-1\right)}$$

Calculate 2 to the power of 3 and get 8.

$$\frac{6\left(\frac{149}{2}+\frac{1}{12}\left(8-2\right)\right)}{10-\left(10^{2}-1\right)}$$

Subtract 2 from 8 to get 6.

$$\frac{6\left(\frac{149}{2}+\frac{1}{12}\times 6\right)}{10-\left(10^{2}-1\right)}$$

Multiply \frac{1}{12} and 6 to get \frac{6}{12}.

$$\frac{6\left(\frac{149}{2}+\frac{6}{12}\right)}{10-\left(10^{2}-1\right)}$$

Reduce the fraction \frac{6}{12} to lowest terms by extracting and canceling out 6.

$$\frac{6\left(\frac{149}{2}+\frac{1}{2}\right)}{10-\left(10^{2}-1\right)}$$

Since \frac{149}{2} and \frac{1}{2} have the same denominator, add them by adding their numerators.

$$\frac{6\times \frac{149+1}{2}}{10-\left(10^{2}-1\right)}$$

Add 149 and 1 to get 150.

$$\frac{6\times \frac{150}{2}}{10-\left(10^{2}-1\right)}$$

Divide 150 by 2 to get 75.

$$\frac{6\times 75}{10-\left(10^{2}-1\right)}$$

Multiply 6 and 75 to get 450.

$$\frac{450}{10-\left(10^{2}-1\right)}$$

Calculate 10 to the power of 2 and get 100.

$$\frac{450}{10-\left(100-1\right)}$$

Subtract 1 from 100 to get 99.

$$\frac{450}{10-99}$$

Subtract 99 from 10 to get -89.

$$\frac{450}{-89}$$

Fraction \frac{450}{-89} can be rewritten as -\frac{450}{89} by extracting the negative sign.

$$-\frac{450}{89}$$

Factor

$$-\frac{450}{89} = -5\frac{5}{89} = -5.056179775280899$$