The answer is -1.

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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 $\boxed {-1}$ .

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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 $\boxed{-1}$ .

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@Vinayak Kumar You've just equated the two expressions, they may not be equal to zero. Also, if a number satisfies the sum of expressions, how can you say that it satisfies both of them individually? Thanks.

Satvik Golechha
- 6 years, 6 months ago

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Can you elaborte what you did after multiplying by 1 or $\omega^{3}$ .

Ayush Choubey
- 6 years, 6 months ago

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We can keep on adding $3$ to the power of $x$ till we don't get $2014$ and $2015$ respectively.

Satvik Golechha
- 6 years, 6 months ago

would u mind solve the function(x^2+x+1=0) again?!!

Rocky Zhang
- 6 years, 6 months ago

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Direct application of the Sri Dharacharya's Rule, aka the Quadratic Formula. You can complete the square, though, if you wish to!

Satvik Golechha
- 6 years, 6 months ago

x^2+x+1 has no rational roots. I think there is a typo in the above post, missing i in the complex roots.

Margarita Melkumova
- 3 years, 6 months ago

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

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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

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