Solve (x+2/2)^3-(x+1)^2+x^2=x^2(x/8+3/4)

$$( \frac { x + 2 } { 2 } ) ^ { 3 } - ( x + 1 ) ^ { 2 } + x ^ { 2 } = x ^ { 2 } ( \frac { x } { 8 } + \frac { 3 } { 4 } )$$

Solve for x

$$x=0$$

Steps for Solving Linear Equation

Multiply both sides of the equation by 8, the least common multiple of 8,4.

$$8\left(\left(\frac{x+2}{2}\right)^{3}-\left(x+1\right)^{2}\right)+8x^{2}=8x^{2}\left(\frac{x}{8}+\frac{3}{4}\right)$$

To raise \frac{x+2}{2} to a power, raise both numerator and denominator to the power and then divide.

$$8\left(\frac{\left(x+2\right)^{3}}{2^{3}}-\left(x+1\right)^{2}\right)+8x^{2}=8x^{2}\left(\frac{x}{8}+\frac{3}{4}\right)$$

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(x+1\right)^{2}.

$$8\left(\frac{\left(x+2\right)^{3}}{2^{3}}-\left(x^{2}+2x+1\right)\right)+8x^{2}=8x^{2}\left(\frac{x}{8}+\frac{3}{4}\right)$$

To find the opposite of x^{2}+2x+1, find the opposite of each term.

$$8\left(\frac{\left(x+2\right)^{3}}{2^{3}}-x^{2}-2x-1\right)+8x^{2}=8x^{2}\left(\frac{x}{8}+\frac{3}{4}\right)$$

To add or subtract expressions, expand them to make their denominators the same. Multiply -x^{2}-2x-1 times \frac{2^{3}}{2^{3}}.

$$8\left(\frac{\left(x+2\right)^{3}}{2^{3}}+\frac{\left(-x^{2}-2x-1\right)\times 2^{3}}{2^{3}}\right)+8x^{2}=8x^{2}\left(\frac{x}{8}+\frac{3}{4}\right)$$

Since \frac{\left(x+2\right)^{3}}{2^{3}} and \frac{\left(-x^{2}-2x-1\right)\times 2^{3}}{2^{3}} have the same denominator, add them by adding their numerators.

$$8\times \frac{\left(x+2\right)^{3}+\left(-x^{2}-2x-1\right)\times 2^{3}}{2^{3}}+8x^{2}=8x^{2}\left(\frac{x}{8}+\frac{3}{4}\right)$$

Do the multiplications in \left(x+2\right)^{3}+\left(-x^{2}-2x-1\right)\times 2^{3}.

$$8\times \frac{x^{3}+3x^{2}\times 2+3x\times 2^{2}+2^{3}-8x^{2}-16x-8}{2^{3}}+8x^{2}=8x^{2}\left(\frac{x}{8}+\frac{3}{4}\right)$$

Combine like terms in x^{3}+3x^{2}\times 2+3x\times 2^{2}+2^{3}-8x^{2}-16x-8.

$$8\times \frac{x^{3}-2x^{2}-4x}{2^{3}}+8x^{2}=8x^{2}\left(\frac{x}{8}+\frac{3}{4}\right)$$

Express 8\times \frac{x^{3}-2x^{2}-4x}{2^{3}} as a single fraction.

$$\frac{8\left(x^{3}-2x^{2}-4x\right)}{2^{3}}+8x^{2}=8x^{2}\left(\frac{x}{8}+\frac{3}{4}\right)$$

To add or subtract expressions, expand them to make their denominators the same. Multiply 8x^{2} times \frac{2^{3}}{2^{3}}.

$$\frac{8\left(x^{3}-2x^{2}-4x\right)}{2^{3}}+\frac{8x^{2}\times 2^{3}}{2^{3}}=8x^{2}\left(\frac{x}{8}+\frac{3}{4}\right)$$

Since \frac{8\left(x^{3}-2x^{2}-4x\right)}{2^{3}} and \frac{8x^{2}\times 2^{3}}{2^{3}} have the same denominator, add them by adding their numerators.

$$\frac{8\left(x^{3}-2x^{2}-4x\right)+8x^{2}\times 2^{3}}{2^{3}}=8x^{2}\left(\frac{x}{8}+\frac{3}{4}\right)$$

Do the multiplications in 8\left(x^{3}-2x^{2}-4x\right)+8x^{2}\times 2^{3}.

$$\frac{8x^{3}-16x^{2}-32x+64x^{2}}{2^{3}}=8x^{2}\left(\frac{x}{8}+\frac{3}{4}\right)$$

Combine like terms in 8x^{3}-16x^{2}-32x+64x^{2}.

$$\frac{8x^{3}+48x^{2}-32x}{2^{3}}=8x^{2}\left(\frac{x}{8}+\frac{3}{4}\right)$$

To add or subtract expressions, expand them to make their denominators the same. Least common multiple of 8 and 4 is 8. Multiply \frac{3}{4} times \frac{2}{2}.

$$\frac{8x^{3}+48x^{2}-32x}{2^{3}}=8x^{2}\left(\frac{x}{8}+\frac{3\times 2}{8}\right)$$

Since \frac{x}{8} and \frac{3\times 2}{8} have the same denominator, add them by adding their numerators.

$$\frac{8x^{3}+48x^{2}-32x}{2^{3}}=8x^{2}\times \frac{x+3\times 2}{8}$$

Do the multiplications in x+3\times 2.

$$\frac{8x^{3}+48x^{2}-32x}{2^{3}}=8x^{2}\times \frac{x+6}{8}$$

Express 8\times \frac{x+6}{8} as a single fraction.

$$\frac{8x^{3}+48x^{2}-32x}{2^{3}}=\frac{8\left(x+6\right)}{8}x^{2}$$

Cancel out 8 and 8.

$$\frac{8x^{3}+48x^{2}-32x}{2^{3}}=\left(x+6\right)x^{2}$$

Use the distributive property to multiply x+6 by x^{2}.

$$\frac{8x^{3}+48x^{2}-32x}{2^{3}}=x^{3}+6x^{2}$$

Calculate 2 to the power of 3 and get 8.

$$\frac{8x^{3}+48x^{2}-32x}{8}=x^{3}+6x^{2}$$

Divide each term of 8x^{3}+48x^{2}-32x by 8 to get -4x+6x^{2}+x^{3}.

$$-4x+6x^{2}+x^{3}=x^{3}+6x^{2}$$

Subtract x^{3} from both sides.

$$-4x+6x^{2}+x^{3}-x^{3}=6x^{2}$$

Combine x^{3} and -x^{3} to get 0.

$$-4x+6x^{2}=6x^{2}$$

Subtract 6x^{2} from both sides.

$$-4x+6x^{2}-6x^{2}=0$$

Combine 6x^{2} and -6x^{2} to get 0.

$$-4x=0$$

Product of two numbers is equal to 0 if at least one of them is 0. Since -4 is not equal to 0, x must be equal to 0.

$$x=0$$