Consider a biquadratic equation 81 x4+216 x3+216 x2+96 x-65=0 whose roots are α, β, γ, δ. Given α, β are real roots and γ, δ are imaginary roots. The value of γ3+δ3-(α+β)3 is equal to

Question

Consider a biquadratic equation 81 x4+216 x3+216 x2+96 x-65=0 whose roots are α, β, &gamma;, δ. Given α, β are real roots and &gamma;, δ are imaginary roots. The value of &gamma;3+δ3-(α+β)3 is equal to (A) (52/9) (B) (53/9) (C) (59/9) (D) (50/9)

Tardigrade Admin · Accepted Answer

81 x4+216 x3+216 x2+96 x-65=0 (3 x x4+4 C1(3 x)3(2)+4 C2(3 x)2(2)2+4 C3(3 x)(2)3+(2)4=81. (3 x+2)4=81 (3 x+2)2=9 text or (3 x+2)2=-9 <img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/mathematics/85228ff03b268411ad3af2d351b1ba75-.png" /> &gamma;3+δ3-(α+β)3 =(&gamma;+δ)(&gamma;2+δ2-&gamma; δ)-(α+β)3 =(-4/3)(((-4/3))2-3 &gamma; δ)-((-4/3))3 =(4/3) ⋅ 3 ⋅ (13/9)=(52/9)

Q. Consider a biquadratic equation $81 x^{4} + 216 x^{3} + 216 x^{2} + 96 x - 65 = 0$ whose roots are $α, β, γ, δ$ . Given $α, β$ are real roots and $γ, δ$ are imaginary roots.
The value of $γ^{3} + δ^{3} - (α + β)^{3}$ is equal to

$\frac{52}{9}$

$\frac{53}{9}$

$\frac{59}{9}$

$\frac{50}{9}$

Q. Consider a biquadratic equation 81x4+216x3+216x2+96x−65=0 whose roots are α,β,γ,δ. Given α,β are real roots and γ,δ are imaginary roots. The value of γ3+δ3−(α+β)3 is equal to

952​

953​

959​

950​

81x4+216x3+216x2+96x−65=0 (3xx4+4C1​(3x)3(2)+4C2​(3x)2(2)2+4C3​(3x)(2)3+(2)4=81 (3x+2)4=81 (3x+2)2=9 or (3x+2)2=−9 γ3+δ3−(α+β)3 =(γ+δ)(γ2+δ2−γδ)−(α+β)3 =3−4​((3−4​)2−3γδ)−(3−4​)3 =34​⋅3⋅913​=952​

Q. Consider a biquadratic equation $81 x^{4} + 216 x^{3} + 216 x^{2} + 96 x - 65 = 0$ whose roots are $α, β, γ, δ$ . Given $α, β$ are real roots and $γ, δ$ are imaginary roots.
The value of $γ^{3} + δ^{3} - (α + β)^{3}$ is equal to

$\frac{52}{9}$

$\frac{53}{9}$

$\frac{59}{9}$

$\frac{50}{9}$