JEE Main 2023 (Online) 8th April Morning ShiftQuestion 1
Let $\alpha,\beta,\gamma$ be the three roots of the equation $x^{3}+bx+c=0$. If $\beta\gamma=1=-\alpha$, then $b^{3}+2c^{3}-3\alpha^{3}-6\beta^{3}-8\gamma^{3}$ is equal to
Quadratic Equation · Higher Degree Polynomial Equations
Solution
Concept. Use the relations between roots and coefficients; $\beta,\gamma$ then satisfy $x^2-x+1=0$, so their cubes are $-1$. Formulas used. (1) For $x^3+bx+c=0$: $\alpha+\beta+\gamma=0$, $\alpha\beta+\beta\gamma+\gamma\alpha=b$, $\alpha\beta\gamma=-c$(2) $x^2-x+1=0\Rightarrow x^3=-1$ (since $x^3+1=(x+1)(x^2-x+1)$)Step 1 - Coefficients. $\alpha=-1$, $\beta\gamma=1$, so $\beta+\gamma=1$. By (1): $$b=\alpha(\beta+\gamma)+\beta\gamma=-1+1=0,\qquad c=-\alpha\beta\gamma=1.$$ Step 2 - Cubes. $\beta,\gamma$ are roots of $x^2-x+1=0$, so by (2) $\beta^3=\gamma^3=-1$; also $\alpha^3=-1$. Step 3 - Value. $$0+2-3(-1)-6(-1)-8(-1)=2+3+6+8=19.$$ Answer: (C)