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Question: If the roots of the cubic equation \(a x ^ { 3 } + b x ^ { 2 } + c x + d = 0\) are in G.P., then....

If the roots of the cubic equation ax3+bx2+cx+d=0a x ^ { 3 } + b x ^ { 2 } + c x + d = 0 are in G.P., then.

A

c3a=b3dc ^ { 3 } a = b ^ { 3 } d

B

ca3=bd3c a ^ { 3 } = b d ^ { 3 }

C

a3b=c3da ^ { 3 } b = c ^ { 3 } d

D

ab3=cd3a b ^ { 3 } = c d ^ { 3 }

Answer

c3a=b3dc ^ { 3 } a = b ^ { 3 } d

Explanation

Solution

Let AR,A,AR\frac { A } { R } , A , A R be the roots of the equation

ax3+bx2+cx+d=0a x ^ { 3 } + b x ^ { 2 } + c x + d = 0

then A3=A ^ { 3 } = Product of the roots =−da= - \frac { d } { a } ⇒\Rightarrow A=−(da)1/3A = - \left( \frac { d } { a } \right) ^ { 1 / 3 }

Since is a root of the equation.

∴aA3+bA2+cA+d=0\therefore a A ^ { 3 } + b A ^ { 2 } + c A + d = 0

⇒\Rightarrow a(−da)+b(−da)2/3+c(−da)1/3+d=0a \left( - \frac { d } { a } \right) + b \left( - \frac { d } { a } \right) ^ { 2 / 3 } + c \left( - \frac { d } { a } \right) ^ { 1 / 3 } + d = 0

⇒ b(da)2/3=c(da)1/3b \left( \frac { d } { a } \right) ^ { 2 / 3 } = c \left( \frac { d } { a } \right) ^ { 1 / 3 } ⇒ b3d2a2=c3dab ^ { 3 } \frac { d ^ { 2 } } { a ^ { 2 } } = c ^ { 3 } \frac { d } { a }⇒ b3d=c3ab ^ { 3 } d = c ^ { 3 } a.