(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 multiply parentheses

3x^2-3x-(x(3x+4)+8)+6+3=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 add all the numbers together, and all the variables

3x^2-3x-(3x^2+4x+8)+9=0

We get rid of parentheses

3x^2-3x^2-3x-4x-8+9=0

We add all the numbers together, and all the variables

-7x+1=0

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

-7x=-1

x=-1/-7

x=1/7

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