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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