यदि f (x) = x3 – 3px2 + qx – 7
∵ मूल समान्तर श्रेणी में हैं।
अतः माना α = a – d, β = a, γ = a + d
∴ α + β + γ = \(\frac{-b}{a}\)
⇒ a – d + a + a + d = \(\frac{3p}{1}\)
⇒ 3a = 3p
⇒ a = P
तथा α·β + β·γ + γ·α = \(\frac{c}{a}\)
(a – d) × a + a × (a + d) + (a + d) (a – d) = \(\frac{q}{a}\)
a = p रखने पर,
(p – d) × p + px (p + d) + (p + d) (p – d) = q
p2 – pd + p2 + pd + p2 – d2 = q
3p2 – d2 = q
3p2 – q = d2
या d2 = 3p2 – q
अब α·β·γ = \(\frac{-d}{a}\)
(a – d) × a × (a + d) = \(\frac{-(-r)}{1}\)
a × (a2 – d2) = r
a = p तथा d2 = 3p2 – q रखने पर,
p × [p2 – (3p2 – q)] = r
p(p2 – 3p + q) = r
p(-2p2 + q) = r
-2p3 + pq = r
Pq – r = 2p3