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Question

(c) If $\alpha, \beta , \gamma $ are the roots of the equation.

${x^3} - p{x^2} + qx - r = 0$

$	herefore {(\alpha + \beta )^{ - 1}} + {(\beta +

Vedclass · Accepted Answer

$4$

Vedclass · Answer

(c) If $\alpha, \beta , \gamma $ are the roots of the equation.

${x^3} - p{x^2} + qx - r = 0$

$	herefore {(\alpha + \beta )^{ - 1}} + {(\beta + \gamma )^{ - 1}} + {(\gamma + \alpha )^{ - 1}} = \frac{{{p^2} + q}}{{pq - r}}$

Given, $p = 0,\,\,q = 4,\,\,\,r = - 1$

==> $\frac{{{p^2} + q}}{{pq - r}} = \frac{{0 + 4}}{{0 + 1}} = 4$.

If $\alpha ,\,\beta ,\,\gamma $ are the roots of the equation ${x^3} + 4x + 1 = 0,$ then ${(\alpha + \beta )^{ - 1}} + {(\beta + \gamma )^{ - 1}} + {(\gamma + \alpha )^{ - 1}} = $

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