Let α = a - d, β = a and y = a + d are the zeros of the given polynomial.
Sum of the zeros

Since α is the zero of the polynomial, therefore f(a) = 0
⇒ f(a) = a3 + 3pa2 + 3qa + r = 0
⇒ a3 + 3pa2 + 3qa + r = 0
On substituting a = - p, we get

⇒ -p3 + 3p3 - 3pq + r = 0
⇒ 2p3 - 3pq + r = 0