If α and β be the roots of x2 + px + q = 0, then ((ωα+ω 2β)(ω2 α +ωβ )/(α2)β + (β2)α is equal to (ω, ω/2) are complex cube roots of unity)

Question

If α and β be the roots of x2 + px + q = 0, then ((ωα+ω 2β)(ω2 α +ωβ )/(α2)β + (β2)α is equal to (ω, ω/2) are complex cube roots of unity) (A) -(q/p) (B) α β (C) -(p/q) (D) ω

Tardigrade Admin · Accepted Answer

Since, α and β are the roots of the equation x2 + px + q = 0, therefore α+β = -p and αβ = q Now, (ωα + ω 2β )(ω 2α + ω β ) = α2+β2 + (ω4+ω2)αβ (∵ ω 3 = 1) = α 2+β 2 - α β . (∵ ω + ω 2 = 1) = (α +β)2 -3α β = p2 - 3q Also, (α2/β) +(β 2/α) = (α 3+β 3/αβ) = ((α +β )3 -3α β (α +β ) /α β ) = (p(3q-p2)/q) ∴ The given expression = ((p2-3q)/(p(3q-p2))q)= -(q/p)

Q. If $α$ and $β$ be the roots of $x^{2} + p x + q = 0$ , then $\frac{( ω α + ω ^{2} β ) ( ω ^{2} α + ω β )}{\frac{α ^{2}}{β} + \frac{β ^{2}}{α}}$ is equal to ( $ω, ω^{2}$ are complex cube roots of unity)

$- \frac{q}{p}$

$α β$

$- \frac{p}{q}$

$ω$

Q. If α and β be the roots of x2+px+q=0, then βα2​+αβ2​(ωα+ω2β)(ω2α+ωβ)​ is equal to (ω,ω2 are complex cube roots of unity)

−pq​

αβ

−qp​

ω

Q. If $α$ and $β$ be the roots of $x^{2} + p x + q = 0$ , then $\frac{( ω α + ω ^{2} β ) ( ω ^{2} α + ω β )}{\frac{α ^{2}}{β} + \frac{β ^{2}}{α}}$ is equal to ( $ω, ω^{2}$ are complex cube roots of unity)

$- \frac{q}{p}$

$α β$

$- \frac{p}{q}$

$ω$