Interscholastic Mathematic League Senior B Division #11 The roots of the equation 3x^3-38x^2+cx-192=0 form a geometric progression. Compute c.
 A: We solve the problem using too much machinery for a contest question that has to be solved quickly!
The following result is useful. Let $r_1$, $r_2$, and $r_3$ be the roots of the cubic equation $x^3+Ax^2+Bx+C=0$. Then (i) $r_1+r_2+r_3=-A$; (ii) $r_1r_2+_2r_3+r_3r_1=B$; and (iii) $r_1r_2r_3=-C$. 
The roots of our cubic form a three-term geometric progression. Say they are $a$, $ar$, and $ar^2$.  We divide the coefficients of our cubic by $3$, to make the coefficient of $x^3$ equal to $1$. So the sum of the roots is $38/3$, and the product of the roots is $192/3=64$.  
Note that the product of the roots is $(a)(ar)(ar^2)$, which is $a^3r^3$. Thus $(ar)^3=64$.  We conclude that $ar=4$. (This assumes that we are dealing with a
geometric progression of real numbers, which is surely what is intended. If $c$ is real, one can prove that $ar$ cannot be one of the complex roots of $z^3=64$.)
The sum of the roots is $38/3$. Thus $a+ar+ar^2=38/3$. Since $ar=4$, we find that $a+ar^2=26/3$. We have $a=4/r$. Substitute in the equation $a+ar^2=26/3$. We obtain $\frac{4}{r} +4r=\frac{26}{3}$.
Divide through by $2$, multiply by $3$, and by $r$, and rearrange. We get
the equation $6r^2-13r+6=0$. By factoring, or otherwise, we find that $r=2/3$ or $r=3/2$. We work with $r=3/2$. (The other choice gives us the same geometric progression, written backwards.) Our geometric progression is therefore $8/3, 4, 6$. 
It follows that $r_1r_2+r_2r_3+r_3r_1=152/3$, and therefore $c=152$.    
Commenst:  To prove the result that connects the coefficients to the roots, note that if the roots of $x^3+Ax^2+Bx+C=0$ are $r_1$, $r_2$, and $r_3$, then 
$x^3+Ax^2+Bx+C$ is the polynomial $(x-r_1)(x-r_2)(x-r_3)$. Expand this product, and compare coefficients.
Much more frequently useful is the fact that if $x^2+Ax+B=0$ has the roots $r_1$ and $r_2$, then $r_1+r_2=-A$ and $r_1r_2=B$.  
There are analogous results for higher degree polynomials.
Added: There is an easy way to solve the problem in a contest situation. Guess that the roots will be rational.  Then by the Rational Roots Theorem, the rational roots are of the form $a/b$, where $a$ and $b$ are integers, $a$ divides $192$, and $b$ is a positive divisor of $3$. The roots cannot all be integers. A little fiddling with possibilities fairly quickly zooms in on the roots. We left this out originally because with a small numerical change, the roots will no longer be rational. The procedure that we used in the main part works in this more general setting, and the ideas are worth knowing.   
