यदि f (x) = x3 + 3px2 + 3qx + r
∵ 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) = \(\frac{3q}{2}\)
a2 – ad + a2 + ad + a2 – d2 = 3q
3a2 – d2 = 3q
a = -p रखने पर,
3(-p)2 – d2 = 3q
-3p2 – d2 = 3q
d2 = 3p2 – 3q
अब α·β·γ = \(\frac{-d}{a}\)
(a – d) × a × (a + d) = \(\frac{-r}{1}\)
a × (a2 – d2) = -r
a = -p तथा d2 = 3p2 – 3q रखने पर,
(-p) x [p2 – (3p2 – 3q)] = -r
-p(p2 – 3p2 + 3q) = -r
p(-2p2 + 3q) = r
-2p3 + 3pq = r
-2p3 + 3pq – r = 0
2p3 – 3pq + r = 0
