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Let, $\alpha$, $\beta$, $\gamma$ be the 3 real roots of the equation, $x^7-3x^4+x=0$, where, $\alpha \gt \beta$, $\alpha \ne 0$, $\beta \ne 0$.

How is, $\alpha$ and $\beta$ related ?

Let, $\alpha=a$,$\beta=b$

The imaginary roots always occurs in pairs, so $\gamma$ must be equal to $0$.

$x^7-3x^4+x$

=$x(x-a)(x-b)(x^4+...)$

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Do you know how to solve a quadratic equation? –  fpqc Jan 14 '13 at 14:10
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closed as too localized by fpqc, Davide Giraudo, Brandon Carter, rschwieb, Andres Caicedo Jan 14 '13 at 16:19

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2 Answers

up vote 4 down vote accepted

Hint: having factored out one power of $x$, let $y=x^3$

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To expand on Ross' hint, $$ x^7-3x^4+x=x(x^6-3x^3+1)=x((x^3)^2-3(x^3)+1) $$ so you know that one root is $0=\gamma$. The other term is a quadratic in $x^3$ so by the quadratic formula $$ x^3=\frac{3\pm\sqrt{5}}{2} $$ so $$ \alpha=\left(\frac{3+\sqrt{5}}{2}\right)^{1/3}\qquad\beta=\left(\frac{3-\sqrt{5}}{2}\right)^{1/3} $$ Now come up with some relations between $\alpha$ and $\beta$. You might look among $\alpha^3, \beta^3$, and $\alpha\beta$, for example.

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