2026/075) The equation $x^3+px+q =0$ where $q\ne 0$ has one root raciprocal of another route , Show that $p+q^2=1$

Let the roots be $m,\frac{1}{m},n$

Using  Vieta's formula we have

$m +\frac{1}{m} + n=0\cdots(1)$

$ m * \frac{1}{m} + m * n + n *\frac{1}{m} =  p \cdots(2)$

We have product of roots $m* \frac{1}{m} *n = - q \cdots(3)$

From(3) $n = -q\cdots(4)$

From (1) $n = - (m + \frac{1}{m})\cdots(5)$ 

From (2)  $1 + n( m + \frac{1}{m}) = p$

Or $1+ n * (-n) = p$ using (5)$

Or $1 - n^2 = p$

Or $1 = p + n^2$ 

Or $1 = p +q^2$ using (4)

Proved