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 (α+β)3-(γ+δ)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 (α+β)3-(&gamma;+δ)3 is equal to (A) (-128/3) (B) 0 (C) (-64/3) (D) (-142/3)

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" /> (α+β)3-(&gamma;+δ)3=((-4/3))3-((-4/3))3=0

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 $(α + β)^{3} - (γ + δ)^{3}$ is equal to

$\frac{- 128}{3}$

0

$\frac{- 64}{3}$

$\frac{- 142}{3}$

$81 x^{4} + 216 x^{3} + 216 x^{2} + 96 x - 65 = 0$
$(3 x x^{4} + 4 C_{1} (3 x)^{3} (2) + 4 C_{2} (3 x)^{2} (2)^{2} + 4 C_{3} (3 x) (2)^{3} + (2)^{4} = 81$
$(3 x + 2)^{4} = 81$
$(3 x + 2)^{2} = 9 or (3 x + 2)^{2} = - 9$

$(α + β)^{3} - (γ + δ)^{3} = (\frac{- 4}{3})^{3} - (\frac{- 4}{3})^{3} = 0$

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 (α+β)3−(γ+δ)3 is equal to

3−128​

0

3−64​

3−142​

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 (α+β)3−(γ+δ)3=(3−4​)3−(3−4​)3=0

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 $(α + β)^{3} - (γ + δ)^{3}$ is equal to

$\frac{- 128}{3}$

$\frac{- 64}{3}$

$\frac{- 142}{3}$

$81 x^{4} + 216 x^{3} + 216 x^{2} + 96 x - 65 = 0$
$(3 x x^{4} + 4 C_{1} (3 x)^{3} (2) + 4 C_{2} (3 x)^{2} (2)^{2} + 4 C_{3} (3 x) (2)^{3} + (2)^{4} = 81$
$(3 x + 2)^{4} = 81$
$(3 x + 2)^{2} = 9 or (3 x + 2)^{2} = - 9$

$(α + β)^{3} - (γ + δ)^{3} = (\frac{- 4}{3})^{3} - (\frac{- 4}{3})^{3} = 0$