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(C) Given the equation $x^3+bx+c=0$ with roots $\alpha, \beta, \gamma$.From the given conditions,$\alpha = -1$ and $\beta \gamma = 1$.Since $\alpha$ i

Vedclass · Accepted Answer

$19$

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(C) Given the equation $x^3+bx+c=0$ with roots $\alpha, \beta, \gamma$.From the given conditions,$\alpha = -1$ and $\beta \gamma = 1$.Since $\alpha$ is a root,it must satisfy the equation: $(-1)^3 + b(-1) + c = 0$,which simplifies to $-1 - b + c = 0$,or $c - b = 1$.From the relation between roots and coefficients,the product of roots $\alpha \beta \gamma = -c$.Substituting $\alpha = -1$ and $\beta \gamma = 1$,we get $(-1)(1) = -c$,which implies $c = 1$.Substituting $c = 1$ into $c - b = 1$,we get $1 - b = 1$,so $b = 0$.The equation becomes $x^3 + 1 = 0$.The roots of $x^3 = -1$ are $-1, -\omega, -\omega^2$,where $\omega$ is the complex cube root of unity.Let $\alpha = -1, \beta = -\omega, \gamma = -\omega^2$.We need to evaluate $b^3 + 2c^3 - 3\alpha^3 - 6\beta^3 - 8\gamma^3$.Substituting the values: $0^3 + 2(1)^3 - 3(-1)^3 - 6(-\omega)^3 - 8(-\omega^2)^3$.$= 0 + 2 - 3(-1) - 6(-\omega^3) - 8(-\omega^6)$.$= 2 + 3 + 6(1) + 8(1) = 2 + 3 + 6 + 8 = 19$.

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 $......$.

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