Solve {a^{2}b^{2}c+[-1/2a^2b^2c-(-3/2a^2b^2c)]}:{-[-(abc)+(-1/8abc)]}

$$\{ a ^ { 2 } b ^ { 2 } c + [ - \frac { 1 } { 2 } a ^ { 2 } b ^ { 2 } c - ( - \frac { 3 } { 2 } a ^ { 2 } b ^ { 2 } c ) ] \} : \{ - [ - ( a b c ) + ( - \frac { 1 } { 8 } a b c ) ] \}$$

Evaluate

$$\frac{16ab}{9}$$

Short Solution Steps

The opposite of -\frac{3}{2}a^{2}b^{2}c is \frac{3}{2}a^{2}b^{2}c.

$$\frac{a^{2}b^{2}c-\frac{1}{2}a^{2}b^{2}c+\frac{3}{2}a^{2}b^{2}c}{-\left(-abc-\frac{1}{8}abc\right)}$$

Combine -\frac{1}{2}a^{2}b^{2}c and \frac{3}{2}a^{2}b^{2}c to get a^{2}b^{2}c.

$$\frac{a^{2}b^{2}c+a^{2}b^{2}c}{-\left(-abc-\frac{1}{8}abc\right)}$$

Combine a^{2}b^{2}c and a^{2}b^{2}c to get 2a^{2}b^{2}c.

$$\frac{2a^{2}b^{2}c}{-\left(-abc-\frac{1}{8}abc\right)}$$

To find the opposite of -abc-\frac{1}{8}abc, find the opposite of each term.

$$\frac{2a^{2}b^{2}c}{-\left(-abc\right)+\frac{1}{8}abc}$$

Factor the expressions that are not already factored.

$$\frac{2ca^{2}b^{2}}{\frac{1}{8}\left(1+8\right)abc}$$

Cancel out abc in both numerator and denominator.

$$\frac{2ab}{\frac{1}{8}\left(1+8\right)}$$

Expand the expression.

$$\frac{2ab}{\frac{9}{8}}$$

Divide 2ab by \frac{9}{8} by multiplying 2ab by the reciprocal of \frac{9}{8}.

$$\frac{2ab\times 8}{9}$$

Multiply 2 and 8 to get 16.

$$\frac{16ab}{9}$$

Expand

$$\frac{16ab}{9}$$

Short Solution Steps

The opposite of -\frac{3}{2}a^{2}b^{2}c is \frac{3}{2}a^{2}b^{2}c.

$$\frac{a^{2}b^{2}c-\frac{1}{2}a^{2}b^{2}c+\frac{3}{2}a^{2}b^{2}c}{-\left(-abc-\frac{1}{8}abc\right)}$$

Combine -\frac{1}{2}a^{2}b^{2}c and \frac{3}{2}a^{2}b^{2}c to get a^{2}b^{2}c.

$$\frac{a^{2}b^{2}c+a^{2}b^{2}c}{-\left(-abc-\frac{1}{8}abc\right)}$$

Combine a^{2}b^{2}c and a^{2}b^{2}c to get 2a^{2}b^{2}c.

$$\frac{2a^{2}b^{2}c}{-\left(-abc-\frac{1}{8}abc\right)}$$

To find the opposite of -abc-\frac{1}{8}abc, find the opposite of each term.

$$\frac{2a^{2}b^{2}c}{-\left(-abc\right)+\frac{1}{8}abc}$$

Factor the expressions that are not already factored.

$$\frac{2ca^{2}b^{2}}{\frac{1}{8}\left(1+8\right)abc}$$

Cancel out abc in both numerator and denominator.

$$\frac{2ab}{\frac{1}{8}\left(1+8\right)}$$

Expand the expression.

$$\frac{2ab}{\frac{9}{8}}$$

Divide 2ab by \frac{9}{8} by multiplying 2ab by the reciprocal of \frac{9}{8}.

$$\frac{2ab\times 8}{9}$$

Multiply 2 and 8 to get 16.

$$\frac{16ab}{9}$$