Find all real values of $x, y$ and $z$ such that $x-\sqrt{yz}=42$, $y-\sqrt{xz}=6,z-\sqrt{xy}=-30$. (COMC) 1997 Part B Question 4 Find all real values of $x, y$ and $z$ such that $x-\sqrt{yz}=42$, $y-\sqrt{xz}=6,z-\sqrt{xy}=-30$.
I was preparing for my COMC Contest so I was doing past years exams. This question really stop me so I want to get some help from others.
I tried substitute $x$ to $a^2$, $y$ to $b^2$ and $z$ to $c^2$. And combining the equations so I can turn this function to 2 variables. But I am stuck because I can't find a way of turning this function to 2 variables. I add the three new equations with a, b and c in pairs, but I can't reduce the equations to 2 variables. I also tried to subtract the equations in pairs, too. I still can't figure it out. Can someone solve my question with the same method too?
Thank you very much!
This question is from Canadian Open Mathematic Challenge (COMC) 1997 Part B Question 4.
 A: Using your substitution and subtracting the equations pairwise, we end up with:
$72 =  42 - (-30) = a^2 - bc - (c^2 - ab ) = (a-c) ( a+b+c)$
$36 = 6 - (-30) = b^2 - ca - (c^2 - ab) = (b-c) (a+b+c )$
$ 36 = 42 - 6 = a^2 - bc - (b^2 - ca) = (a-b) ( a + b + c) $
Hence $ a-c = 2 (b-c) = 2(a-b)$, or that $ b = \frac{ a+c}{2}$.
Substituting this into the second equation gives us $(a-c)^2 = 24 $ so $|a-c| = 2 \sqrt{6} $.
Substituting this into the first equation gives us $84 = 2a^2 - ac - c^2 = (a-c)(2a + c) $, so $ |2a+c| = 7 \sqrt{6} $, where $a-c, 2a+c$ have the same sign.
Hence, the solutions are $(a, b, c) = \pm ( 3 \sqrt{6}, 2 \sqrt{6}, \sqrt{6} )$, which gives us $ (x,y,z) = (54, 24, 6 )$.

Notes:

*

*Given the symmetry in the equations, taking the pairwise difference and (attempting to) factoring out $a-b$ is a common approach. It is often harder to work with the pairwise sum.

A: Here's another approach you may find useful. First of course set $x=a^2, y=b^2, z=c^2$ and the system reads
\begin{align}
&a^2-bc=42\\
&b^2-ac=6\\
&c^2-ab=-30\\
\end{align}
Now take the linear combination $(1)c+(2)a+(3)b$. We notice that the sum of the LHS is zero and hence we have found a simpler linear relation
$$42c+6a-30b=0$$
Taking a different linear combination $b(1)+c(2)+a(3)$ gives zero again when we add the LHS and therefore we get another linear relation.
$$42b+6c-30a=0$$
Solving the last two linear equations for $a,b$ in terms of $c$ we obtain the extremely simple relations
$$b=2c~,~ a=3c$$
and now substituting back into our original system we obtain $c^2=6$. All in all, the solutions are
$$(a,b,c)=\{(3\sqrt{6}, 2\sqrt{6}, \sqrt{6}),-(3\sqrt{6}, 2\sqrt{6}, \sqrt{6})\}$$
and for the solution of the original system, the second one cannot create a solution, while the first one generates the only solution
$$(x,y,z)=(54,24,6)$$
In such systems we see that it can be useful to find derivative quantities that are simpler than the ones contained in the original system. This system of equations has a high degree of symmetry on the LHS, so one should take advantage of that.
