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Given equation:
$x^3 + px + q = 0$
Since $\alpha, \beta, \gamma$ are the roots of this equation, we have:
$\alpha + \beta + \gamma = 0$
$\alpha\beta + \beta\gamma + \gamma\alpha = p$
$\alpha\beta\gamma = -q$

Now, the roots of the required equation are:
$y_1 = \frac{\alpha+\beta}{\gamma^2} = \frac{-\gamma}{\gamma^2} = -\frac{1}{\gamma}$
$y_2 = \frac{\beta+\gamma}{\alpha^2} = \frac{-\alpha}{\alpha^2} = -\frac{1}{\alpha}$
$y_3 = \frac{\alpha+\gamma}{\beta^2} = \frac{-\beta}{\beta^2} = -\frac{1}{\beta}$

Let $y = -\frac{1}{x} \implies x = -\frac{1}{y}$.

Substituting $x = -\frac{1}{y}$ into the original equation $x^3 + px + q = 0$:
$\left(-\frac{1}{y}\right)^3 + p\left(-\frac{1}{y}\right) + q = 0$
$-\frac{1}{y^3} - \frac{p}{y} + q = 0$

Multiplying throughout by $-y^3$:
$1 + py^2 - qy^3 = 0$
$qy^3 - py^2 - 1 = 0$

Therefore, the required cubic equation is:
$qx^3 - px^2 - 1 = 0$