Matematické Fórum

stuart clark · 27. 03. 2011 15:37

If $\alpha,\beta,\gamma$ be the roots of the equation $x^3+x-1=0$ .Then Calculate value of $\frac{\beta}{\alpha}+\frac{\gamma}{\beta}+\frac{\alpha}{\gamma}=$

musixx · 28. 03. 2011 12:23 — Editoval musixx (28. 03. 2011 12:56)

In general, those tasks are used to lead to so called symmetric polynomials, and Viete's formulas (consult e.g. Wikipedia for details).

But it is not this case, since there is no symmetric polynomial in the numerator of a fraction $\frac{\beta^2\gamma+\gamma^2\alpha+\alpha^2\beta}{\alpha\beta\gamma}$ , while there is a symmetric polynomial in its denominator.

Moreover, your number is not well-defined, since it depends on an order of the roots.

Example 1, an illustrative one:

$(x-1)(x-2)(x-3)$ , and $\frac12+\frac23+\frac31=4\frac16\neq3\frac56=\frac13+\frac32+\frac21$ .

Example 2: it could be shown that your number is either $\frac{-3+{\rm i}\sqrt{31}}2$ or $\frac{-3-{\rm i}\sqrt{31}}2$ .

Pavel · 28. 03. 2011 13:52 — Editoval Pavel (28. 03. 2011 13:56)

↑ stuart clark:

Remark: Similar problem can be found at http://forum.matweb.cz/viewtopic.php?id=21735

Matematické Fórum

#1 27. 03. 2011 15:37

cubic equation

#2 28. 03. 2011 12:23 — Editoval musixx (28. 03. 2011 12:56)

Re: cubic equation

#3 28. 03. 2011 13:52 — Editoval Pavel (28. 03. 2011 13:56)

Re: cubic equation

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