Sol. α + β = -b/a, αβ = c/a
Roots if the equation a3 x3 + abcx + c3 = 0 are
x = -abc ± √(abc)2 – 4a3 c3/2a3
= (-b/a) (c/a) ± √(b/a)2 (c/a)2 – 4(c/a)3 / 2
(α + β) (α β) ± √( α + β)2 (αβ)2 – 4(αβ)3 / 2
= (α β) ((α + β) ± √( α – β)2 / 2
= (α β) ((α + β) ± (α – β)/2 = α2 β, α β2
Let γ and δ be the required roots. Then
γ = α2 β and δ = α β2.
ALTERNATE SOLUTION :
ax2 + bx + c = 0 has roots α and β. (given)
⇒ α + β = -b/a and α β = c/a
Now, a3 x2 + abcx + c3 = 0
Divides the equation by c2, we get
a3/c2 x2 + abcx/c2 + c3/c2 = 0, a(ax/c)2 + b(ax/c) + c = 0
⇒ ax/c = α, β are the roots
⇒ x = c/a α, c/a β are the roots
⇒ x = α β α , α β β are the roots
⇒ x α2 β, α β2 are the roots
ALTERNATE SOLUTION :
Divide the equation by a3, we get
x2 + b/a. c/a. x + (c/a)3 = 0
⇒ x2 – (α + β) (α β)x + (α β)3 = 0
⇒ x2 – α2 βx - α β2 x + (α β)3 = 0
⇒ x (x – α2 β) - α β2 (x – α2 β) = 0
⇒ (x – α2 β) (x - α β2) = 0
⇒ x = α2 β, α β2 which is the required answer.
