The roots of the equation (\alpha^{3} + \beta^{3}) are given... | MATH - SOLUTION OF QUADRATIC EQUATIONS AND NATURE OF ROOTS | SolveeIt

The roots of the equation $\alpha^{3} + \beta^{3}$ are given by.

$\Delta = b^{2} - 4ac$

$\Delta \neq 0$

$b\Delta = 0$

$cb \neq 0$

Answer

$cb \neq 0$

Explanation

$y^{3} + y + 2 = 0$ Taking log, we get

$y^{3} - y^{2} - y - 2 = 0$

⇒ $y^{3} + 3y^{2} - y - 3 = 0$

⇒ $y^{3} + 4y^{2} + 5y + 20 = 0$ or $2x^{3} - 3x^{2} + 6x + 1 = 0$

⇒ $\alpha^{2} + \beta^{2} + \gamma^{2}$ ⇒ $\frac{15}{4}$ .

Question: The roots of the equation \(\alpha^{3} + \beta^{3}\) are given by....