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TimbaLanD

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[math]

x = 3 \sqrt{(x+3)^3 - (x+1)^3 + (x+8)^3)}

[/math]

Now. that one's easy! Square both side and multiply out the cubes to get a cubic equation- then use the cubic formula.

 

The way Atheist interpreted it (you didn't have enough parentheses to make it clear what the square root applied to), squaring gives a 6th degree polynomial equation- and there is no general formula for the solution of 6th degree equations.

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Ok, so if I understood correctly, the equation you are trying to solve is

[math]

x = \left[(x+3)^3 - (x+1)^3 + (x+8)^3 \right]^{1/3}.

[/math]

As said, take this to its third power, leading to

[math]

x^3 = (x+3)^3 - (x+1)^3 + (x+8)^3.

[/math]

Multiply that out. As yourdadonapogos said, the x³-terms will cancel, leading to an equation of the type ax²+bx+c=0 which then can be solved using the pq-equation.

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So its:

[math]

x = 3 \sqrt{(x+3)^3 - (x+1)^3 + (x+8)^3)}

[/math]

 

I separated the equation to make it e little easier, so for those three parts I got :

1.

[math](x+3)^3=(x+3)^{2}(x+3)=(x^{2}+6x+9)(x+3)[/math]

[math]=x^{3}+3x^{2}+6x^{2}+18x+27= x^{3} + 9x^{2} + 18x + 27 [/math]

 

2.

[math](x+1)^3=(x+1)^{2}(x+1)=(x^{2}+2x + 1)(x+1)[/math]

[math]=x^{3}+x^{2}+2x^{2}+2x+x+1=x^{3} + 3x^{2} + 3x + 1[/math]

 

3.

[math](x+8)^{3}=(x+8)^{2}(x+8)=(x^{2}+16x+64)(x+8)=[/math]

[math]x^{3}+8x^{2}+16x^{2}+128x+64x+512=x^{3} 24x^{2} + 192x + 512[/math]

 

And after we do the operations inside the root we get:

[math]x^{3} + 30x^{2} + 207x + 538[/math]

 

So if I'm correct the equation now is:

 

[math]x^{3}=x^{3} + 30x^{2} + 207x +538[/math]

 

As we have the same [math]x^{3}[/math] on both sides, then we eleminate it, and then it looks like this:

 

[math]30x^{2} + 207x +538[/math]

 

And the rest is easy!

 

I hope I haven't made any stupid (forgotten something) mistake!

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