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(B) Given that the cubic equation has real coefficients and one root is $\alpha = 3 + 4i$.Since complex roots occur in conjugate pairs for equations w

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$-150$

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(B) Given that the cubic equation has real coefficients and one root is $\alpha = 3 + 4i$.Since complex roots occur in conjugate pairs for equations with real coefficients,the second root must be $\beta = 3 - 4i$.Let the cubic equation be $x^3 + ax^2 + bx + c = 0$. Since the coefficient of $x^2$ is zero,we have $a = 0$.According to Vieta's formulas,the sum of the roots is $\alpha + \beta + \gamma = -a = 0$.Substituting the known roots: $(3 + 4i) + (3 - 4i) + \gamma = 0$.$6 + \gamma = 0$,which gives $\gamma = -6$.The product of the roots is $\alpha \beta \gamma = (3 + 4i)(3 - 4i)(-6)$.Calculating the product of the conjugate pair: $(3 + 4i)(3 - 4i) = 3^2 - (4i)^2 = 9 + 16 = 25$.Therefore,$\alpha \beta \gamma = 25 	imes (-6) = -150$.

In a cubic equation,the coefficient of $x^2$ is zero 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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