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
Hello, kahandasa!
A very strange problem . . .
$\displaystyle \text{Solve for }x+y+z.$
. . $\displaystyle \begin{array}{c}19x^3+95x+57\:=\:0 \\ 19y^3+95y+57\:=\:0 \\ 19z^3+95z+57\:=\:0 \end{array}$
$\displaystyle x,y\text{ and }z\text{ are not equal to each other.}$
$\displaystyle \text{First of all, the equations can be simplified to: }\:\begin{array}{c}x^3 + 5x + 3 \:=\:0 \\ y^3 + 5y + 3 \:=\:0 \\ z^3 + 5z + 3 \:=\:0\end{array}$
$\displaystyle \text{Evidently, we have the three distinct roots }x,\,y,\,z$
. . $\displaystyle \text{of a cubic equation: }\,u^3 + 0u^2 + 5u + 3 \:=\:0$
$\displaystyle \text{Therefore: }\:x+y+z \:=\:0$
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:
$\displaystyle (u-x)(u-y)(u-z)=u^3-(x+y+z)u^2+... = u^3+ 0\times u^2+5u+3$
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