Solve -128/1*1/n^3(2n^3+3n^2+n)/6

$$\frac { - 128 } { 1 } \cdot \frac { 1 } { n ^ { 3 } } \frac { ( 2 n ^ { 3 } + 3 n ^ { 2 } + n ) } { 6 }$$

Evaluate

$$-\frac{128}{3}-\frac{64}{n}-\frac{64}{3n^{2}}$$

Short Solution Steps

Anything divided by one gives itself.

$$-128\times \frac{1}{n^{3}}\times \frac{2n^{3}+3n^{2}+n}{6}$$

Express -128\times \frac{1}{n^{3}} as a single fraction.

$$\frac{-128}{n^{3}}\times \frac{2n^{3}+3n^{2}+n}{6}$$

Multiply \frac{-128}{n^{3}} times \frac{2n^{3}+3n^{2}+n}{6} by multiplying numerator times numerator and denominator times denominator.

$$\frac{-128\left(2n^{3}+3n^{2}+n\right)}{n^{3}\times 6}$$

Cancel out 2 in both numerator and denominator.

$$\frac{-64\left(2n^{3}+3n^{2}+n\right)}{3n^{3}}$$

Factor the expressions that are not already factored.

$$\frac{-64n\left(n+1\right)\left(2n+1\right)}{3n^{3}}$$

Cancel out n in both numerator and denominator.

$$\frac{-64\left(n+1\right)\left(2n+1\right)}{3n^{2}}$$

Expand the expression.

$$\frac{-128n^{2}-192n-64}{3n^{2}}$$

Expand

$$-\frac{128}{3}-\frac{64}{n}-\frac{64}{3n^{2}}$$

Short Solution Steps

Anything divided by one gives itself.

$$-128\times \frac{1}{n^{3}}\times \frac{2n^{3}+3n^{2}+n}{6}$$

Express -128\times \frac{1}{n^{3}} as a single fraction.

$$\frac{-128}{n^{3}}\times \frac{2n^{3}+3n^{2}+n}{6}$$

Multiply \frac{-128}{n^{3}} times \frac{2n^{3}+3n^{2}+n}{6} by multiplying numerator times numerator and denominator times denominator.

$$\frac{-128\left(2n^{3}+3n^{2}+n\right)}{n^{3}\times 6}$$

Cancel out 2 in both numerator and denominator.

$$\frac{-64\left(2n^{3}+3n^{2}+n\right)}{3n^{3}}$$

Factor the expressions that are not already factored.

$$\frac{-64n\left(n+1\right)\left(2n+1\right)}{3n^{3}}$$

Cancel out n in both numerator and denominator.

$$\frac{-64\left(n+1\right)\left(2n+1\right)}{3n^{2}}$$

Expand the expression.

$$\frac{-128n^{2}-192n-64}{3n^{2}}$$