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(B) Since the coefficients are real,the complex roots must occur in conjugate pairs. Given $\alpha = 3 + 4i$,the second root is $\beta = 3 - 4i$.Let t

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$-150$

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(B) Since the coefficients are real,the complex roots must occur in conjugate pairs. Given $\alpha = 3 + 4i$,the second root is $\beta = 3 - 4i$.Let the cubic equation be $x^3 + ax^2 + bx + c = 0$. Since the coefficient of $x^2$ is $0$,the sum of the roots is $\alpha + \beta + \gamma = 0$.Substituting the known values: $(3 + 4i) + (3 - 4i) + \gamma = 0$.$6 + \gamma = 0 \implies \gamma = -6$.The product of the roots is $\alpha \beta \gamma = (3 + 4i)(3 - 4i)(-6)$.Since $(3 + 4i)(3 - 4i) = 3^2 + 4^2 = 9 + 16 = 25$,we have $\alpha \beta \gamma = 25 	imes (-6) = -150$.

In a cubic equation,the coefficient of $x^2$ is $0$ and the remaining coefficients are real. If one root is $\alpha = 3 + 4i$,and the remaining roots are $\beta$ and $\gamma$,then find the value of $\alpha \beta \gamma$.

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