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By rational root theorem, the only possible rational roots of the above polynomial could be $\pm1$ which are not in this case. Therefore, the roots are Irrational of Imaginary.

Either it is not a RMO question or my approach is not complete.

this is surely from RMO (jharkhand RMO 2015)...but i have solved this problem integrating 3 concepts : theory of equations, inequalities and then final proof by contradiction

@Gyanendra Prakash
–
well i approached like this:-
let the roots be a,b,c
a+b+c=3
ab+bc+ac=0
abc=1
assuming roots to be rational..i took a=p1/q1,b as p2/q2 and c as p3/q3

so i got p1/q1+p2/q2+p3/q3=3---------eq.1
p1p2/q1q2 + p2p3/q2/q3 + p1p3/q1q3 = 0-----eq.2
and p1p2p3=q1q2q3 -----------eq-3

procceding with equation 1

i got after expanding and replacing q1q2q3 by p1p2p3...

recoiprocal of eq.2=3 (after three steps of monotonous algebraic expansion)

taking eq2 as x+y+z = 0 and then its reciprocal from above as 1/x+x/y+1/z = 3

by A.M-G.M we know that
(x+y+z)(1/x+1/y+1/z)>=9

but here we are getting it as 3*0=0

therfore by contradiction roots can't be rational...

@Gyanendra Prakash
–
I don't know, but if it is a RMO Question it should be solved somewhat the way you did it. I can't actually understand your solution, should use LaTex and you left the most part of your solution that is the steps with which you solved.

In a group of seven persons every person is asked to write the sum of ages of all the other six persons.if all the seven sums form a six element set {63,64,65,66,68,70}..then find the individual ages assuming ages are whole numbers..

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TopNewestone question was: prove that the roots of the equation x^3-3x^2-1 = 0 are never rational.

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By rational root theorem, the only possible rational roots of the above polynomial could be $\pm1$ which are not in this case. Therefore, the roots are Irrational of Imaginary.

Either it is not a RMO question or my approach is not complete.

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this is surely from RMO (jharkhand RMO 2015)...but i have solved this problem integrating 3 concepts : theory of equations, inequalities and then final proof by contradiction

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so i got p1/q1+p2/q2+p3/q3=3---------eq.1 p1p2/q1q2 + p2p3/q2/q3 + p1p3/q1q3 = 0-----eq.2 and p1p2p3=q1q2q3 -----------eq-3

procceding with equation 1

i got after expanding and replacing q1q2q3 by p1p2p3...

recoiprocal of eq.2=3 (after three steps of monotonous algebraic expansion)

taking eq2 as x+y+z = 0 and then its reciprocal from above as 1/x+x/y+1/z = 3

by A.M-G.M we know that (x+y+z)(1/x+1/y+1/z)>=9

but here we are getting it as 3*0=0

therfore by contradiction roots can't be rational...

i suppose it's corrcet..

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In a group of seven persons every person is asked to write the sum of ages of all the other six persons.if all the seven sums form a six element set {63,64,65,66,68,70}..then find the individual ages assuming ages are whole numbers..

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find all triplets such that (a,b,c) of positive integers such that a<=b<=c: 3 does not divide b.c and (1+1/a)(2+1/b)(3+1/c)=13

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last time cut off was 45. by the way how much are you getting this time?

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If the roots of x^3+px^2+qx+r=0 are all real and +ve...prove pq-9r>=0 and q^3-27r^2>=0

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