If $\alpha$, $\beta$, $\gamma$ are the zeros of $x^3 + 4x + 1$, then calculate the value of: $$(\alpha + \beta)–1 + (\beta + \gamma)–1 + (\gamma + \alpha)–1$$
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Is it a contest problem? – Inceptio Apr 07 '13 at 17:26
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1Do you mean $4^x$, as in the title, or $4x$, as in the question? Since you say polynomial, I assume the latter. – robjohn Apr 07 '13 at 17:29
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2I get the feeling that the expression should be reciprocals, otherwise the problem is fairly trivial (it doesn't seem to be too much harder however) – Andrew D Apr 07 '13 at 17:32
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The poster must have made a TeX error; it must be reciprocals. So, put the expression as a single fraction, and use the fact that symmetric rational functions of the roots are rational functions of the coefficients. Multiply out $(x - \alpha) (x - \beta) (x - \gamma)$ to se how. – Eric Jablow Apr 07 '13 at 21:55
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HINT: You want to calculates $2(\alpha+\beta+\gamma)-3$. There is a formula to find $\alpha+\beta+\gamma$ when you know the coeficients of the polynomial.
diff_math
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Note that $(\alpha + \beta)-1+(\beta+\gamma)-1+(\gamma +\alpha)-1=\left(2\alpha+2\beta+2\gamma\right)-3=2(\alpha+\beta+\gamma)-3$. We know from Vieta's formulae that:
$$\alpha+\beta+\gamma=-\frac{0}{1}=0$$
Therefore, $(α + β)–1 + (β + γ)–1 + (γ + α)–1=2(0)-3=-3$.
Thomas Russell
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Hint:
Sum of the roots of equation : $x^3+bx^2+cx+d=0$ is $-b$.
So convert it into this form. You see that co-efficient of $x^2$ is zero, so the expression's value is $-3$.
Hail Vieta!!
Inceptio
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Hint $\ $ Exploit symmetry. The sum $\rm = 2(a+b+c) - 3,\:$ and $\rm\:a+b+c = $ -coef of $\rm\,x^2\,$ by Vieta.
Remark $\ $ Generally one can calculate any symmetric polynomial of the roots in the same way, since, by a simple algorithm of Gauss, any symmetric polynomial may be written as a polynomial in the elementary symmetric polynomials (Fundamental Theorem of Symmetric Polynomials)
