If α and β are the roots of ax2 + bx + c = 0, then α3 + β3 =A) \(

1.	If α and β are the roots of ax2 + bx + c = 0, then α3 + β3 =A) \(\cfrac{3abc-b^3}{a^3}\)B) \(\cfrac{3abc-b^3}{c^3}\)C) \(\cfrac{b^2-3abc}{a^3}\)D) \(\cfrac{b^2-3abc}{c^3}\)
Answer» Correct option is (A) \(\frac{3abc-b^3}{a^3}\) \(\alpha+\beta=\frac{-b}a\) and \(\alpha\beta=\frac ca\) \(\because\) \((\alpha+\beta)^3\) \(=\alpha^3+\beta^3+3\alpha\beta\,(\alpha+\beta)\) \(=(\frac{-b}a)^3-3.\frac ca.\frac{-b}a\) \(=\frac{-b^3}{a^3}+\frac{3bc}{a^2}\) \(=\frac{3abc-b^3}{a^3}\) Correct option is A) \(\cfrac{3abc-b^3}{a^3}\)

If α and β are the roots of ax2 + bx + c = 0, then α3 + β3 =A) \(\cfrac{3abc-b^3}{a^3}\)B) \(\cfrac{3abc-b^3}{c^3}\)C) \(\cfrac{b^2-3abc}{a^3}\)D) \(\cfrac{b^2-3abc}{c^3}\)

Answer»

Correct option is (A) \(\frac{3abc-b^3}{a^3}\)

\(\alpha+\beta=\frac{-b}a\) and \(\alpha\beta=\frac ca\)

\(\because\) \((\alpha+\beta)^3\) \(=\alpha^3+\beta^3+3\alpha\beta\,(\alpha+\beta)\)

\(=(\frac{-b}a)^3-3.\frac ca.\frac{-b}a\)

\(=\frac{-b^3}{a^3}+\frac{3bc}{a^2}\) \(=\frac{3abc-b^3}{a^3}\)

Correct option is A) \(\cfrac{3abc-b^3}{a^3}\)