If - - are the roots ax2-2bx-c-0- then -3-3 - -2-3 - -3-2 -

The correct option is C c^2a^3(c+2b) As α ,β are roots of ax^ 2 2bx+c=0 α +β = 2b a and αβ = c a Therefore, α ^ 3 β ^ 3 +α ^ ...

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Standard XII

Mathematics

Common Roots

If α, β are...

Question

If $α, β$ are the roots $a x^{2} - 2 b x + c = 0$ , then $α^{3} β^{3} + α^{2} β^{3} + α^{3} β^{2} =$

\frac{c^{2}}{a^{3}} (c + 2 b)

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\frac{b c^{3}}{a^{3}}

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\frac{c^{3}}{a^{3}} (c - 2 b)

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\frac{b c}{a^{3}}

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Solution

The correct option is C $\frac{c^{2}}{a^{3}} (c + 2 b)$
As $α, β$ are roots of $a x^{2} - 2 b x + c = 0$
$α + β = \frac{2 b}{a}$ and $α β = \frac{c}{a}$
Therefore,
$α^{3} β^{3} + α^{2} β^{3} + α^{3} β^{2} = {(α β)}^{3} + {(α β)}^{2} (α + β) = {(\frac{c}{a})}^{2} + (\frac{c}{a}) (\frac{2 b}{a}) = \frac{c^{2}}{a^{3}} (c + 2 b)$
Hence, option 'A' is correct .

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Similar questions

Q. prove that

a^{3} {(b - c)}^{3} + b^{3} {(c - a)}^{3} + c^{3} {(a - b)}^{3} =

a^{3} (c - b) + b^{3} (a - c) + c^{3} (b - a)

Q. Lets consider quadratic equation

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\frac{b_{2}}{b_{1}} = \frac{b_{3}}{b_{2}} = \frac{b_{4}}{b_{3}} =

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Now if

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c_{2}

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\frac{1}{c_{2}} - \frac{1}{c_{1}} = \frac{1}{c_{3}} - \frac{1}{c_{2}} = \frac{1}{c_{4}} - \frac{1}{c_{3}} =

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If α,β are the roots ax2−2bx+c=0, then α3β3+α2β3+α3β2=

The correct option is C c2a3(c+2b)As α,β are roots of ax2−2bx+c=0α+β=2ba and αβ=caTherefore,α3β3+α2β3+α3β2=(αβ)3+(αβ)2(α+β)=(ca)2+(ca)(2ba)=c2a3(c+2b)Hence, option 'A' is correct .

If $α, β$ are the roots $a x^{2} - 2 b x + c = 0$ , then $α^{3} β^{3} + α^{2} β^{3} + α^{3} β^{2} =$