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Question
If α, β and γ are the roots of the cubic equation x3 + 2x2 + 3x + 4 = 0, form a cubic equation whose roots are `1/alpha, 1/beta, 1/γ`
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Solution
The new roots are `1/alpha, 1/beta, 1/γ`
`sum1 = 1/alpha + 1/beta + 1/γ`
= `(betaγ + alphaγ + alphabeta)/(alpha beta γ)`
= `(- 3)/4`
`sum2 = 1/(alphabeta) + 1/(betaγ) + 1/(γalpha)`
= `(alpha + beta + γ)/(alpha beta γ)`
= `(- 2)/(- 4)`
= `1/2`
`sum3 = 1/(alpha γ)`
= `- 1/4`
Required equation is
`x^3 - (- 3/4)x^2 + 1/2x - (- 1/4)` = 0
`x^3 + 3/4 x^2 + x/2 + 1/4` = 0
⇒ 4x3 + 3x2 + 2x + 1 = 0
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