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Mathematics Question on Quadratic Equations

If α and β\alpha \, and \, \beta are the roots of the equation ax+bx+c=0, then the value of α3 + β3\alpha^{3} \, + \, \beta^{3} is

A

3abc + b3a3\frac{3abc \, + \, b^{3}}{a^3}

B

a3 + b33abc\frac{a^3 \, + \, b^3}{3abc}

C

3abc − b3a3\frac{3abc \, - \, b^3}{a^3}

D

−(3abc + b3)a3\frac{-(3abc \, + \, b^3)}{a^3}

Answer

3abc − b3a3\frac{3abc \, - \, b^3}{a^3}

Explanation

Solution

Given : α\alpha & β\beta are roots of equation
ax + bx + c = 0
∴  α +β =−ba & αβ = ca\therefore \, \, \alpha \, + \beta \, = -\frac{b}{a} \, \& \, \alpha \beta \, = \, \frac{c}{a}
Now, α3+β3 =(α +β)3− 3αβ(α+β)\alpha^3 + \beta^3 \, =(\alpha \, + \beta)^3 - \, 3\alpha\beta(\alpha+\beta)
⇒ α3+β3=(−ba)3 −3ca(−ba)\Rightarrow \, \alpha^3+\beta^3 = \bigg(-\frac{b}{a}\bigg)^3 \, -3 \frac{c}{a} \bigg(-\frac{b}{a}\bigg)
⇒  α3+β3 = −b2a3+3bca2\Rightarrow \, \, \alpha^3 + \beta^3 \, = \, -\frac{b^2}{a^3} + \frac{3bc}{a^2}
⇒   α3+β3 = −b3 + 3abca3\Rightarrow \, \, \, \alpha^3 + \beta^3 \, = \, \frac{-b^3 \, + \, 3abc}{a^3}