Solve [(5/6x^{2}+3x)^{3}-(5/6x^2-3x)^3]:(2x^3)-[(5/2x-8)^2+4(10x+1)]

$$[ ( \frac { 5 } { 6 } x ^ { 2 } + 3 x ) ^ { 3 } - ( \frac { 5 } { 6 } x ^ { 2 } - 3 x ) ^ { 3 } ] : ( 2 x ^ { 3 } ) - [ ( \frac { 5 } { 2 } x - 8 ) ^ { 2 } + 4 ( 10 x + 1 ) ]$$

Factor

$$-41$$

Solution Steps

Factor the expressions that are not already factored in \frac{\left(\frac{5}{6}x^{2}+3x\right)^{3}-\left(\frac{5}{6}x^{2}-3x\right)^{3}}{2x^{3}}.

$$factor(\frac{\frac{1}{216}\times 108\left(25x^{2}+108\right)x^{3}}{2x^{3}}-\left(\left(\frac{5}{2}x-8\right)^{2}+4\left(10x+1\right)\right))$$

Cancel out 2x^{3} in both numerator and denominator.

$$factor(\frac{1}{216}\times 54\left(25x^{2}+108\right)-\left(\left(\frac{5}{2}x-8\right)^{2}+4\left(10x+1\right)\right))$$

Expand the expression.

$$factor(\frac{25}{4}x^{2}+27-\left(\left(\frac{5}{2}x-8\right)^{2}+4\left(10x+1\right)\right))$$

Use the distributive property to multiply 4 by 10x+1.

$$factor(\frac{25}{4}x^{2}+27-\left(\left(\frac{5}{2}x-8\right)^{2}+40x+4\right))$$

To find the opposite of \left(\frac{5}{2}x-8\right)^{2}+40x+4, find the opposite of each term.

$$factor(\frac{25}{4}x^{2}+27-\left(\frac{5}{2}x-8\right)^{2}-40x-4)$$

Subtract 4 from 27 to get 23.

$$factor(\frac{25}{4}x^{2}+23-\left(\frac{5}{2}x-8\right)^{2}-40x)$$

Factor out \frac{1}{4}.

$$\frac{25x^{2}+92-\left(5x-16\right)^{2}-160x}{4}$$

Simplify.

$$-41$$

Evaluate

$$-41$$