2 Equations, 1 Variable. Still Unsolvable?

Note : Here roots of Cubics are
$-1\quad ,\quad \omega \quad ,\quad { \omega }^{ 2 }\quad \quad$ .

out of them only $\omega \quad ,\quad { \omega }^{ 2 }$ . satisfy the 2nd equation given in question. !!

Since $1\quad +\quad \omega \quad +\quad { \omega }^{ 2 }\quad \quad =\quad 0$ .

Hence Simultaneous solutions are : $\omega \quad ,\quad { \omega }^{ 2 }$ .

Sum is

$\quad \omega \quad +\quad { \omega }^{ 2 }\quad \quad =\quad -1$ .

Q.E.D

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here : $\quad \omega \quad ,\quad { \omega }^{ 2 }\quad \quad \longrightarrow \quad Cube\quad roots\quad of\quad unity$ .

I think the best solution is by Vinayak Kumar. Because the roots satisfy the two equations are those satisfy the combined equation: $x^{2015} + x^{2014} + x^3 + 2x^2 + 2x + 2 = 0$ And by Vieta's we get the answer as $\boxed {-1}$ .

I actually solved it and hence proved it is -1 by a long way. We find that the first equation can be factorized into: $x^3 + 2x^2 + 2x + 1 = (x+1)(x^2 + x + 1)$ and hence the three roots are $-1, -\frac {1}{2} \pm i \frac {\sqrt{3}}{2}$ .

We see that $x = -1$ is not a solution of the second equation $x^{2015} + x^{2014} + 1 = 0$ but the two complex roots $-\frac {1}{2} \pm i \frac {\sqrt{3}}{2}$ are. Therefore, the sum of roots is $\boxed {-1}$ .

We use the transitive property to get $x^{2015}+x^{2014}+1=x^3+2x^2+2x+1$ . This simplifies to $x^{2015}+x^{2014}+1-x^3-2x^2-2x-1=0$ .

Then by Vieta's we cleary see the sum is $\boxed{-1}$ .

Fortunately, we can find the $3$ roots of the cubic. $x^3+2x^2+2x+1=(x+1)(x^2+x+1)$ So, the roots are $-1$ , $\frac{-1+\sqrt{3}}{2}$ and $\frac{-1-\sqrt{3}}{2}$ . Let's call any one of them $\omega$ . By observation, the other one is automatically $\omega^2$ . Also, by Vieta's Formula, we have the following relations: $\omega^3=1$ and $1+\omega+\omega^2=0$ In the second equation which we got, we can multiply any term with $\omega^3=1$ , which won't change anything. So, finally, if $k \equiv \mod3$ , and $l \equiv 2\mod3$ , then, $x^k+x^l+1=0$ Woah!! This is our second equation. $x^{2015}+x^{2014}+1=0$ . So, there are only two roots common to both the equations. Adding them up, $\omega+\omega^2$ , is thus, $\boxed{-1}$

roots of first equation: (i) by hit and trial we find the first root as -1; therefore x=-1 hence x+1=0 now (x3 + 2x2 + 2x +1)/(x+1) =x2+x+1 solving x2+x+1 we get the root by sri dharacharya method we get roots x= w,w2

(ii)roots of the equation are w and w2 therefore the common roots are w and w2

so w+w2=-1