Question

α,β and ¥ are roots of cubic polynomial ax^3+bx^2+cx+d then 1/αβ+1/β¥+1/¥α​

Answer 1

EXPLANATION.

α,β,γ are the roots of the polynomial,

⇒ F(x) = ax³ + bx² + cx + d.

As we know that,

Sum of zeroes of the cubic equation,

⇒ α + β + γ = -b/a.

Products of zeroes of cubic polynomial two at a time,

⇒ αβ + βγ + γα = c/a.

Products of zeroes of cubic polynomial,

⇒ αβγ = -d/a.

⇒ 1/αβ + 1/βγ + 1/γα.

⇒ (βγ)(γα) + (αβ)(γα) + (αβ)(βγ)/(αβ)(βγ)(γα).

⇒ (αβγ²) + (βγα²) + (αγβ²)/(α²β²γ²).

⇒ αβγ[γ + α + β]/(αβγ)².

⇒ [α + β + γ]/(αβγ).

⇒ -b/a/-d/a

⇒ -b/a X a/-d.

⇒ b/d.

                                                                                                                                       

MORE INFORMATION.

Nature of the factors of the quadratic expression.

(1) = Real and different, if b² - 4ac > 0.

(2) = Rational and different, if b² - 4ac is a perfect square.

(3) = Real and equal, if b² - 4ac = 0.

(4) = If D < 0 Roots are imaginary and unequal or complex conjugate.

Answer 2

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$\huge {AηsωeR} ✍$

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$\sf \:  ⟼Since, \: \alpha , \beta , \gamma \: are \: the \: roots \: of \: the$

$\sf \:  polynomial \: f(x) = {ax}^{3} + {bx}^{2} + cx + d, \: then$

$\begin{gathered}\begin{gathered}\bf \begin{cases} &\sf{ \alpha + \beta + \gamma = - \dfrac{b}{a} } \\ &\sf{ \alpha \beta \gamma = - \dfrac{d}{a} } \end{cases}\end{gathered}\end{gathered}$

☆ Now, Consider

$\sf \:  ⟼ \: \dfrac{1}{ \alpha \beta } + \dfrac{1}{ \beta \gamma } + \dfrac{1}{ \gamma \alpha }$

☆ On taking LCM, we get

$\sf \:  ⟼ = \dfrac{ \ \gamma + \alpha + \beta }{ \alpha \beta \gamma }$

$\sf \:  ⟼ \: = \: \dfrac{ \alpha + \beta + \gamma }{ \alpha \beta \gamma }$

$\sf \:  ⟼ \: = \dfrac{ - \dfrac{b}{a} }{ - \dfrac{d}{a} }$

$\sf \:  ⟼ \: = \dfrac{b}{d}$

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