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Find the equation whose roots are the fourth powers of the roots of the equation $x^3 + x + 1 = 0$. Hence find the sum of the fourth powers of the roots of the equation $x^3 + x + 1 = 0$

I am really stuck. The answer at the back says cubic is $u^3 - 2u^2 + 5u -1 = 0$ and sum of roots is 2

Rubicon
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m.bazza
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  • One approach: If $x$ satisfies $x^3 + x + 1 = 0$, then $x^4$ satisfies $(x^4)^{3/4} + (x^4)^{1/4} + 1 = 0$. The only problem is that this is not a polynomial equation... but you can fix that. – Mr. Chip Apr 26 '17 at 08:57

2 Answers2

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So we seek a cubic in $y = x^4, $ where $x^3+x+1=0$.

Hence we have $y = x^4 = x(x^3) = -x(x+1) = -x^2-x$, using which, $y^2 = (x^4)^2 = x^2(x+1)^2 = x^4+2x^3+x^2 = -3x-2$.

Similarly, $y^3 = (x^4)^3 = -x^3(x+1)^3 = 5x^2-x-3$.

Now it is easy to eliminate (the powers of) $x$ and get $y^3-2y^2 +5y = 1\dots$

Macavity
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Let $\alpha$ be a root of the polynomial $x^3+x+1$. Then $1,\alpha,\alpha^2$ is a $\mathbb{Q}$-basis of the number field $\mathbb{Q}(\alpha)$ (you can use Gauss' lemma to show that the polynomial is irreducible). Write $\beta = \alpha^4$ and write $1,\beta,\beta^2,\beta^3 \in \mathbb{Q}(\alpha)$ with respect to the mentioned basis. Using linear algebra, you can then get a $\mathbb{Q}$-linear relation between $1,\beta,\beta^2,\beta^3$, which gives you a non-zero polynomial in $\mathbb{Q}[x]$ of which $\beta$ is a zero.

The sums of the roots of a polynomial $x^n + a_{n-1}x^{n-1} +...$ equals $-a_{n-1}$, which follows from expanding the factorization of the polynomial.

Ovidius
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