If one root of the equation x^2-px+q=0 is the square of the other, show that p^3-q(3p+1)-q^2=0. Provided q is not equal to 1

Answer 1

let the roots be m and n
where n = m^2

we know sum of roots = m+n = p
and the product of the roots is mn = q

LS = p^3 - q(3p+1) - q^2
= (m+n)^3 - mn(3m+3n+1) - m^2n^2
= m^3 + 3m^2n + 3mn^2 + n^3 - 3m^2 n - 3m n^2 - mn - m^2 n^2
= m^3 + n^3 - mn - m^2n^2
but n = m^2, so ...

= m^3 + m^6 - m^3 -m^6
= 0
= RS , as required