(x2+2)^3-(x-1)^3=x(3x+4)+8

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Solution for (x2+2)^3-(x-1)^3=x(3x+4)+8 equation: 



(x2+2)^3-(x-1)^3=x(3x+4)+8

We move all terms to the left:

(x2+2)^3-(x-1)^3-(x(3x+4)+8)=0

We add all the numbers together, and all the variables

(+x^2+2)^3-(x-1)^3-(x(3x+4)+8)=0



We calculate terms in parentheses: -(x(3x+4)+8), so:

x(3x+4)+8

We multiply parentheses

3x^2+4x+8

Back to the equation:

-(3x^2+4x+8)



We get rid of parentheses

(+x^2+2)^3-3x^2-(x-1)^3-4x-8=0

We add all the numbers together, and all the variables

-3x^2+(+x^2+2)^3-4x-(x-1)^3-8=0

We move all terms containing x to the left, all other terms to the right

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

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