Let $α, β, γ$ be roots of $x^{3} + Px + q = 0$ then

see full answer

Your Exam Success, Personally Taken Care Of

1:1 expert mentors customize learning to your strength and weaknesses – so you score higher in school , IIT JEE and NEET entrance exams.

An Intiative by Sri Chaitanya

An equation whose roots $α^{3}, β^{3}, γ^{3} is x^{3} + 3 {qx}^{2} + (3 q^{2} + p^{3}) x + q^{3} = 0$

$α^{3} + β^{3} + γ^{3} = 3 αβγ$

An equation whose roots are $α^{3} + β^{3}, β^{3} + γ^{3}, γ^{3} + α^{3} is x^{3} + 6 {qx}^{2} + (P^{3} + 12 q^{2}) x + 3 P^{3} q + 8 q^{3} = 0$

$α^{4} + β^{4} + γ^{4} = 2 P^{2}$

answer is A, B, C, D.

(Unlock A.I Detailed Solution for FREE)

Best Courses for You

JEE

NEET

Foundation JEE

Foundation NEET

CBSE

Detailed Solution

$\begin{array}{l} y = α^{3} \Rightarrow α = y^{1 / 3} \\ ∴ y + {Py}^{1 / 3} + q = 0 \end{array}$
$\begin{array}{l} (y + q)^{3} = - P^{3} y \\ y^{3} + 3 {qy}^{2} + (3 q^{2} + p^{3}) y + q^{3} = 0 \\ α^{3} + β^{3} + γ^{3} = - 3 q = 3 αβγ \\ Let y = β^{3} + γ^{3} = - 3 q - α^{3} \end{array}$
As $α^{3}$ is a root of equation
$\begin{array}{l} (- 3 q - y)^{3} + 3 q (- 3 q - y)^{2} + (3 q^{2} + p^{3}) (- 3 q - y) + q^{3} = 0 \\ y^{3} + 6 {qy}^{2} + (p^{3} + 12 q^{2}) y + 3 P^{3} q + 8 q^{3} = 0 \\ α^{2} + β^{2} + γ^{2} = {(α + β + γ)}^{2} - 2 (βγ + γα + αβ) \\ \Rightarrow α^{2} + β^{2} + γ^{2} = 0 - 2 p, \\ α^{2} β^{2} + β^{2} γ^{2} + γ^{2} α^{2} = (αβ + βγ + γα)^{2} - 2 αβγ (α + β + γ) = P^{2} \\ α^{4} + β^{4} + γ^{4} = {(α^{2} + β^{2} + γ^{2})}^{2} - 2 (β^{2} γ^{2} + γ^{2} α^{2} + α^{2} β^{2}) \\ α^{4} + β^{4} + γ^{4} = {(- 2 p)}^{2} - 2 (p^{2}) = 2 p^{2} \end{array}$