If alpha, beta and gamma are the roots of x^3 | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) Given the cubic equation $x^3 + px + q = 0$. Since $\alpha, \beta, \gamma$ are the roots,by Vieta's formulas: $\alpha + \beta + \gamma = 0$ (coeff

Vedclass · Accepted Answer

$-3q$

Vedclass · Answer

(A) Given the cubic equation $x^3 + px + q = 0$. Since $\alpha, \beta, \gamma$ are the roots,by Vieta's formulas: $\alpha + \beta + \gamma = 0$ (coefficient of $x^2$ is $0$). $\alpha\beta + \beta\gamma + \gamma\alpha = p$. $\alpha\beta\gamma = -q$. We know the algebraic identity: $\alpha^3 + \beta^3 + \gamma^3 - 3\alpha\beta\gamma = (\alpha + \beta + \gamma)(\alpha^2 + \beta^2 + \gamma^2 - \alpha\beta - \beta\gamma - \gamma\alpha)$. Since $\alpha + \beta + \gamma = 0$,the right side of the identity becomes $0$. Therefore,$\alpha^3 + \beta^3 + \gamma^3 = 3\alpha\beta\gamma$. Substituting $\alpha\beta\gamma = -q$,we get $\alpha^3 + \beta^3 + \gamma^3 = 3(-q) = -3q$.

If $\alpha, \beta$ and $\gamma$ are the roots of $x^3 + px + q = 0$,then the value of $\alpha^3 + \beta^3 + \gamma^3$ is equal to

Similar Questions

If for a positive integer $n$,the quadratic equation $x(x + 1) + (x + 1)(x + 2) + \dots + (x + n - 1)(x + n) = 10n$ has two consecutive integral solutions,then $n$ is equal to:

The equation $\sqrt{3 x^{2}+x+5}=x-3$,where $x$ is real,has

Two real numbers $\alpha$ and $\beta$ are such that $\alpha + \beta = 3$ and $|\alpha - \beta| = 4$. Then $\alpha$ and $\beta$ are the roots of the quadratic equation:

For the equation $3x^2 + px + 3 = 0, p > 0$,if one root is the square of the other,then $p$ is equal to:

If the roots of the equation $bx^2 + cx + a = 0$ are imaginary,then for all real values of $x$,the expression $3b^2x^2 + 6bcx + 2c^2$ is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module