Hello... I am Adnan from Bosnia... I am preparing for exam to be accepted in college ...

I need help for this problem...

$\displaystyle x+y+z=1;

x^2+y^2+z^2=1;

x^3+y^3+z^3=1;$

Solve..

Thanks

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- Jul 6th 2009, 04:39 PMc3nar1usFast help needed
Hello... I am Adnan from Bosnia... I am preparing for exam to be accepted in college ...

I need help for this problem...

$\displaystyle x+y+z=1;

x^2+y^2+z^2=1;

x^3+y^3+z^3=1;$

Solve..

Thanks - Jul 6th 2009, 05:57 PMKasper
Because we can see that an increase in degree does not change the value of the answer, we know that the variables MUST be something that always stays the same with changing whole number powers. Which could be $\displaystyle \pm1$ or $\displaystyle 0$. But because a) the answer is $\displaystyle 1$ (So it is obvious the variables cannot be zero), and b) the answer is ALWAYS 1, we know that the variables must be positive. (If they were negative they would switch between negative and positive between the square and cube functions, which they don't).

Using the above information, we can see that 1 variable must be 1, while the other two variables must be 0. - Jul 7th 2009, 02:56 AMc3nar1us
- Jul 7th 2009, 03:11 AMsimplependulum
Suppose $\displaystyle x,y,z ~~$ are the roots of a cubic equation

we have $\displaystyle x+y+z = 1$ so the equation is $\displaystyle

u^3 - u^2 + bu + c = 0 $

At the same time ,

$\displaystyle x^2 + y^2 + z^2 = (x+y+z)^2 - 2(xy+yz+zx) = 1$ so

$\displaystyle 1 - 2(xy+yz+zx) = 1 \implies xy+yz+zx = 0$

The equation :$\displaystyle u^3 - u^2 + c = 0 $

And

$\displaystyle x^3 = x^3 - x^2 + c + ( x^2 - x + \frac{c}{x}) - c + x - \frac{c}{x} = 0+0 - c + x - \frac{c}{x} = x - c - \frac{c}{x}$

So $\displaystyle x^3 + y^3 + z^3 = x+y+z - c-c-c - c (\frac{1}{x} + \frac{1}{y} + \frac{1}{z}) = 1 - 3c - c [ \frac{ xy+yz+zx}{xyz}] = 1-3c $

It is given that $\displaystyle x^3 + y^3 + z^3 = 1$

Therefore , $\displaystyle 1 = 1-3c \implies c = 0$

The equation is u^3 - u^2 = 0

u = 0 , 0, 1

Therefore , (x,y,z) = (0,0,1),(0,1,0),(1,0,0)

If the values of the right hand sides are not 1,the above method can be applied to solve it , It is not just used to prove the set of the solution is (0,0,1) in this case . - Jul 7th 2009, 03:16 AMc3nar1us