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Question

(A) Given the quadratic equation $Ax^2 + Bx + C = 0$.From the relation between roots and coefficients,we have $\alpha + \beta = -\frac{B}{A}$ and $\al

Vedclass · Accepted Answer

$\frac{3ABC - B^3}{A^3}$

Vedclass · Answer

(A) Given the quadratic equation $Ax^2 + Bx + C = 0$.From the relation between roots and coefficients,we have $\alpha + \beta = -\frac{B}{A}$ and $\alpha \beta = \frac{C}{A}$.We know the algebraic identity $\alpha^3 + \beta^3 = (\alpha + \beta)^3 - 3\alpha \beta (\alpha + \beta)$.Substituting the values,we get $\alpha^3 + \beta^3 = (-\frac{B}{A})^3 - 3(\frac{C}{A})(-\frac{B}{A})$.$\alpha^3 + \beta^3 = -\frac{B^3}{A^3} + \frac{3BC}{A^2}$.Taking $A^3$ as the common denominator,$\alpha^3 + \beta^3 = \frac{-B^3 + 3ABC}{A^3} = \frac{3ABC - B^3}{A^3}$.

If $\alpha$ and $\beta$ are roots of the equation $Ax^2 + Bx + C = 0$,then the value of $\alpha^3 + \beta^3$ is

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