If the roots of the equation x^3+3px^2+3qx-8= | vedclass

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(D) Let the roots of the cubic equation $x^3+3px^2+3qx-8=0$ be $\frac{a}{r}, a, ar$. From the relation between roots and coefficients,the product of t

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-$8$

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(D) Let the roots of the cubic equation $x^3+3px^2+3qx-8=0$ be $\frac{a}{r}, a, ar$. From the relation between roots and coefficients,the product of the roots is $\frac{a}{r} 	imes a 	imes ar = -(-8) = 8$. Thus,$a^3 = 8$,which implies $a = 2$. Since $a=2$ is a root,it must satisfy the equation: $(2)^3 + 3p(2)^2 + 3q(2) - 8 = 0$. $8 + 12p + 6q - 8 = 0$,which simplifies to $12p + 6q = 0$,or $q = -2p$. Substituting $q = -2p$ into the expression $\frac{q^3}{p^3}$,we get $\frac{(-2p)^3}{p^3} = \frac{-8p^3}{p^3} = -8$.

If the roots of the equation $x^3+3px^2+3qx-8=0$ are in a geometric progression,then $\frac{q^3}{p^3}=$

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