dont tell me...

[TimbaLanD]

Dont tell me the answer but how easy is this to do?

 

X = 3 √(X+3)3 - (X+1)3 + (X+8) 3

 

find X...

[timo]

you mean [math]x = 3 \sqrt{(x+3)^3} - (x+1)^3 + (x+8)^3 [/math]? I'm not even convinced that it can be analytically solved at all.

[TimbaLanD]

I am not sure how to go about this one... you think it can't be solved?

 

[math]

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

[/math]

[Country Boy]

[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.

[Mr Skeptic]

Just to make clarify, should that be a cube root, or 3 times a square root?

[TimbaLanD]

cube root...

[ydoaPs]

Once you cube both sides and multiply everything out, you can simplify it to a quadratic equation. From there, it's just a matter of the quadratic formula.

[TimbaLanD]

now i need help! Anyone?

[timo]

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.

[TimbaLanD]

I got

 

[math]

x^3 = -x^3+30x^2-168x+538

[/math]

 

Is that right?

[insane_alien]

you've got a sign wrong somewhere the x^3's should cancel.

[thedarkshade]

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!

[the tree]

Hmmm, I got [math]0=30\,{x}^{2}+216\,x+538[/math] at the last step.

 

For slightly easier (and more accurate) expansions: [math](a+b)^3 = a^3 +3a^2 b+3ab^2 +b^3[/math], or more generally for any natural power you can use the binomial expansion.

[thedarkshade]

Hmmm, I got [math]0=30\,{x}^{2}+216\,x+538[/math] at the last step.

Gives imaginary solutions, just like mine!

[ydoaPs]

Hmmm, I got [math]0=30\,{x}^{2}+216\,x+538[/math] at the last step.

 

For slightly easier (and more accurate) expansions: [math](a+b)^2 = a^3 +3a^2 b+3ab^2 +b^3[/math], or more generally for any natural power you can use the binomial expansion.

I was under the impression (a+b)²=a²+2ab+b².

[the tree]

Bah, a typo, you know what I meant.

 

 

[math]

(a+b)^3 = a^3 +3a^2 b+3ab^2 +b^3

[/math]

[insane_alien]

I was under the impression (a+b)²=a²+2ab+b².

 

me too. think tree made a whoopsie.

 

EDIT: ninja posted by the tree himself how embarassing.

[TimbaLanD]

So what is X?

[ydoaPs]

Why don't you tell us? Solve for zero then use the quadratic formula.

[TimbaLanD]

Is x an integer?

[ydoaPs]

No.

[the tree]

Not in the slightest.

[5614]

At the risk of giving the game away I will ask: are you familiar with complex (imaginary) numbers?

[TimbaLanD]

Do you mean the "i" and "j".. I need to refresh my A/L Maths! Is the answer going to be an imaginarty number? I am lost!

[the tree]

No, it'll be complex (have both real and imaginary parts).