If α, β, γ are roots of x3-5 x+4=0, then (α3+β3+γ3)2 is equal to

Question

If α, β, &gamma; are roots of x3-5 x+4=0, then (α3+β3+&gamma;3)2 is equal to (A) 12 (B) 13 (C) 169 (D) 144

Tardigrade Admin · Accepted Answer

Given, the roots of x3-5 x+4 =0 are α, β text and &gamma; ∴ α+β+&gamma; =0 α β+β &gamma;+&gamma; α =5 and α β &gamma;=-4 Since, α+β+&gamma; =0 ∴ α3+β3+&gamma;3 =3 α β &gamma; =3 ×(-4)=-12 Hence, (α3+β3+&gamma;3)2 =(-12)2=144

Q. If $α, β, γ$ are roots of $x^{3} - 5 x + 4 = 0$ , then $(α^{3} + β^{3} + γ^{3})^{2}$ is equal to

12

13

169

144

Given, the roots of
$x^{3} - 5 x + 4 = 0$ are $α, β and γ$
$∴ α + β + γ = 0$
$α β + β γ + γ α = 5$ and $α β γ = - 4$
Since, $α + β + γ = 0$
$∴ α^{3} + β^{3} + γ^{3} = 3 α β γ$
$= 3 \times (- 4) = - 12$
Hence, $(α^{3} + β^{3} + γ^{3})^{2} = (- 12)^{2} = 144$

Q. If α,β,γ are roots of x3−5x+4=0, then (α3+β3+γ3)2 is equal to

12

13

169

144

Given, the roots of x3−5x+4=0 are α,β and γ ∴α+β+γ=0 αβ+βγ+γα=5 and αβγ=−4 Since, α+β+γ=0 ∴α3+β3+γ3=3αβγ =3×(−4)=−12 Hence, (α3+β3+γ3)2=(−12)2=144

Q. If $α, β, γ$ are roots of $x^{3} - 5 x + 4 = 0$ , then $(α^{3} + β^{3} + γ^{3})^{2}$ is equal to

Given, the roots of
$x^{3} - 5 x + 4 = 0$ are $α, β and γ$
$∴ α + β + γ = 0$
$α β + β γ + γ α = 5$ and $α β γ = - 4$
Since, $α + β + γ = 0$
$∴ α^{3} + β^{3} + γ^{3} = 3 α β γ$
$= 3 \times (- 4) = - 12$
Hence, $(α^{3} + β^{3} + γ^{3})^{2} = (- 12)^{2} = 144$