Given that α and ß are roots of quadratic equation ax2 + bx + c = 0
\(\therefore\) sum of roots = -b/a
i.e., α + ß = -b/a----(1)
And product of roots = c/a
i.e., αß = c/a---(2)
Now, (aα + b)-3 + (aß + b)-3
= (aα - (aα + aß))-3 + (aß - (aα + aß))-3
(From (1), -b/a = α + ß ⇒ b = -(aα + aß))
= (-aß)-3 + (-aα)-3
= -a-3ß-3 - a-3α-3
= \(-\frac1{a^3}(\frac1{α^3}+\frac1{ß^3})\)
\(=-\frac1{\alpha^3}(\frac{\alpha^3+\beta^3}{(\alpha\beta)^3})\)
\(=-\frac1{a^3}(\frac{(\alpha+\beta)(\alpha^2+\beta^2-\alpha\beta)}{(\alpha\beta)^3})\)
\(=-\frac1{a^3}(\frac{(\alpha+\beta)((\alpha+\beta)^2-3\alpha\beta)}{(\alpha\beta)^3})\)
\(=-\frac1{a^3}\left(\cfrac{-\frac ba((\frac{-b}a)^2-\frac{3c}a)}{(\frac ca)^3}\right)\) (From (1) and (2))
\(=-\frac1{c^3}\left(-\frac ba(\frac{b^2-3ac}{a^2})\right)\) = \(\frac{b}{a^3c^3}(b^2-3ac)\)
= \(\frac{b^3}{a^3c^3}-\frac{3abc}{a^3c^3}\)
