2011/084) If p q r are roots of equation x^3 – 3px^2 + 3q^2x – r^3 = 0 then prove that p = q = r

proof:

as sum of roots is 3p so p+ q + r = 3p

so there exists t such that

q = p- t, r = p+ t

coefficient of x = pq + pr + qr = 3q

or p(q+r) + qr = 3q^2

or p* 2p + (p – t)(p+t) = 3(p-t)^2

or 3p^2 – t^2 = 3p^2-6pt + 3t^2

or 4t^2-6pt = 0 or

t(2t-3p) = 0 ..1

constant = p(p+t)(p-t) = (p+t)^3

p+t = 0 or p(p-t) = (p+t)^2

or p^2-pt = p^2 +2 pt + t^2

or t^2 + 3pt =0

t(t+3p) = 0 ….2

fom (1) and (2) t = 0

or

2t-3p = 0 and t+ 3p = 0 which again gives t = 0 (also p = 0 this is redundant as a sub set of 1^st solution)

So t = 0

So p = q =r

hence proved