1. ## equations help

solve for x+y+z if
19x^3+95x+57=0
19y^3+95y+57=0
19z^3+95z+57=0
x,yand z are not equal to each other

2. ## Re: equations help

Originally Posted by kahandasa
solve for x+y+z if
19x^3+95x+57=0
19y^3+95y+57=0
19z^3+95z+57=0
All of the equations are identical, so x = y = z. Just solve the first equation for x (if possible) and then you will have y and z as well.

3. ## Re: equations help

ya but it is very diificullt. the only way is to play with the eqation and obtain x+y+z

4. ## Re: equations help

Hello, kahandasa!

A very strange problem . . .

$\text{Solve for }x+y+z.$

. . $\begin{array}{c}19x^3+95x+57\:=\:0 \\ 19y^3+95y+57\:=\:0 \\ 19z^3+95z+57\:=\:0 \end{array}$

$x,y\text{ and }z\text{ are not equal to each other.}$

$\text{First of all, the equations can be simplified to: }\:\begin{array}{c}x^3 + 5x + 3 \:=\:0 \\ y^3 + 5y + 3 \:=\:0 \\ z^3 + 5z + 3 \:=\:0\end{array}$

$\text{Evidently, we have the three distinct roots }x,\,y,\,z$
. . $\text{of a cubic equation: }\,u^3 + 0u^2 + 5u + 3 \:=\:0$

$\text{Therefore: }\:x+y+z \:=\:0$

5. ## Re: equations help

i dont understand why x+y+z=0

6. ## Re: equations help

Originally Posted by kahandasa
i dont understand why x+y+z=0
Viete's relations: the sum of the roots of an n-th degree polynomial equation is -1 times the coefficient of the (n-1)-th degree term divided by the coefficient of the n-th degree term. Since there is no term in the square of the variable and we have a cubic the sum of the roots is zero.

In this case we can be explicit, with soroban's notation:

$(u-x)(u-y)(u-z)=u^3-(x+y+z)u^2+... = u^3+ 0\times u^2+5u+3$

etc..

Also all the roots are different since by Descartes' rule of signs there are zero positive roots and one negative root, hence exactly one real root, and as this is a polynomial with real coefficients the remaining two complex roots are a conjugate pair and hence not equal.

CB