6yup the ans is -1/2.
$If $\alpha,\beta$ and $\gamma$ be the Roots of the equation $x^3+x-1=0$.Then Calculate\\\\ value of $\frac{\beta}{\alpha}+\frac{\gamma}{\beta}+\frac{\alpha}{\gamma}=$
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1just an attempt :
\sum{\frac{\alpha}{\beta}}=\frac{\sum \alpha^2\beta}{\alpha\beta\gamma}\\ \\ \left(\alpha + \beta +\gamma \right)\left(\alpha\beta +\beta\gamma +\gamma\alpha \right)=6\sum \alpha^2\beta +3 \alpha\beta\gamma \\ \sum \alpha^2\beta=-\frac{1}{2} \\ \boxed{\sum{\frac{\alpha}{\beta}}=-\frac{1}{2} }
1i agree with you KUNL,
expanding,(α+β+γ)(αβ+βγ+αγ)...i m getting only a part of expression required in the numerator of (β/α+γ/β+α/γ)
1@learner,yes that is,but what about the numerator?i m getting that by expanding (α+β+γ)(αβ+βγ+αγ) with some more additionals terms.plzz help!
1well first things first question is wrong this is what i felt at first sight[3][3]....but keeping in view man posted it i thought some hitch might be there....but i don't find any way other than to find roots explicitly...if there is man plz post the solution!
1let the roots be a b c for simplicity.let us find the equation whose roots are
b/a,c/b,a/c.
let y=b/a=1/a2c as abc=1
but 1=a3+a as a is a root.
y=a3+a/a2c=a+1/ac. but 1/ac=b
y=a+b=-c as a+b+c=0
c=-y
therefore the equation is y3+y+1=0
therefore b/a+c/b+a/c=0
try out a similar problem
x3-6x2+3x+1=0
p q r are the roots find the value of p2q+q2r+r2p