Question

If ${\alpha} {\beta} {\gamma}$ are the zeros of the polynomial $f(x)=x^{3}+px^{2}+qx-r$,then $\frac{1}{\alpha \beta}+ \frac{1}{\beta \gamma} +\frac{1}{\gamma \alpha}$=

Answer 1

SOLUTION :

Given :  α, β, γ are the three Zeroes of the cubic  polynomial f(x) = x³ - px² + qx - r

On comparing with ax³ + bx² + cx + d,

a = 1 , b= - p ,c = q , d =  - r

Sum of zeroes of cubic  polynomial= −coefficient of x² / coefficient of x³

α + β + γ = −(-p /1

α + β + γ = p ………………….(1)

Product of zeroes of cubic  polynomial = - constant term / coefficient of x³

αβγ = - d/a  

αβγ = - (-r)/1

αβγ = r ……………….(2)

The value of : 1/αβ + 1/βγ + 1/γα  

=  γ + α + β / αβγ

= α + β + γ/ αβγ

= p / r  

[From eq 1 & 2]

1/αβ + 1/βγ + 1/γα = p/r

Hence, the value of 1/αβ + 1/βγ + 1/γα is p/r.

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Answer 2

Given:

α,β,γ are zero of polynomial

$f(x)=x^{3}+px^{2}+qx-r$

To find:

$\frac{1}{\alpha \beta}+ \frac{1}{\beta \gamma} +\frac{1}{\gamma \alpha}$

Solution :

We have α,β,γ are zero of the polynomial

f(x) = x³-px²+qx-r

Then

α+β+γ = -(-p)/1 = P

αβ+βγ+

1