Problem 1683

Suppose the common roots are $a, b$ and the third root of $g$ is $c$. Using Vieta's formulas, or comparing coefficients in
$$g(x) = x^{3} - \gamma x - 6 = (x - a)(x - b)(x - c) = x^{3} - (a+b+c)x^{2} + (ab+bc+ca)x - abc$$
give $a + b + c = 0$ and $abc = 6$. Hence, $-c = a + b$ and so
$$ab(a + b) = -6. \qquad (15)$$
The roots $a, b$ are the roots of the quadratic polynomial $g - f$:
$$g(x) - f(x) = 3x^2 + (\gamma - \alpha)x - (\beta + 6) = 3(x - a)(x - b) \quad (16)$$
hence
$$\alpha - \gamma = 3(a + b) \qquad (17)$$
$$\beta + 6 = -3ab.$$

To prove the equation $(\alpha - \gamma)^3 + 9\gamma(\alpha - \gamma) + 162 = 0$, we use (17) to write this in equivalent form as
$$(a + b)^3 + (a + b)\gamma + 6 = 0 \quad \text{or}$$
$$a^3 + a\gamma + b^3 + b\gamma + 3ab(a + b) + 6 = 0.$$
Because $g(a) = g(b) = 0$, we have $a^3 + \gamma a = b^3 + \gamma b = 6$. Using (15) we see now that the above equation indeed holds true. Finally, from (18) we obtain
$$54 = (\alpha - \gamma)(\beta + 6) = (\beta + 6)(\alpha - \gamma - 9) + 9(\beta + 6)$$
which simplifies to
$$-(\beta + 6)(\alpha - \gamma - 9) = 9\beta.$$
Using this and multiplying the expression for $g(x) - f(x)$ in (16) by $\alpha - \gamma - 9$, we obtain
$$3(\alpha - \gamma - 9)x^2 + (\alpha - \gamma - 9)(\gamma - \alpha)x + 9\beta = 0$$
which has $a$ and $b$ as its roots, as required.