Find $x,y,z$ for the given conditions $$4x^2+25y^2+9z^2-10xy-15yz-6zx=0$$
$$x+y+z=5$$
I tried two approaches 
1) Substituting $x$ as $5-y-z$ in the first equation but didn't work out, I was getting $39y^2+19z^2-31yz-90y-70z+100=0$ which can't be factorized
2) First equation corresponds to $a^2+b^2+c^2-ab-bc-ca=0$, which means $a^3+b^3+c^3=3abc$, but didn't get a breakthrough.
I am stuck here, please help me.
 A: $$(a-b)^2+(b-c)^2+(c-a)^2=2(?)=0$$
Now for real $d,d^2\ge0$
If $a,b,c$ are real, we can establish $$a-b=b-c=c-a=0$$
A: Just wanted to see. Here is how it looks if you make the Hessian matrix congruent to a diagonal matrix, the one that gives
$$  \frac{1}{4} \left(4x - 5 y  - 3 z \right)^2 + \frac{3}{4} \left( 5y  - 3 z\right)^2 = 4x^2 + 25 y^2 + 9 z^2 - 15 yz - 6 zx - 10 xy   $$
is $ Q^T D Q = H  $
The theorem is that the form does factor, allowing complex coefficients, if and only if the discriminant is zero. In this case, we see that $\det H = 0.$
As you saw, setting to zero gives us $5y=3z,$ after which $4x = 5y + 5y = 10y,$ giving $2x=5y=3z$ 
=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=
$$ P^T H P = D  $$
$$\left( 
\begin{array}{rrr} 
1 & 0 & 0 \\ 
 \frac{ 5 }{ 4 }  & 1 & 0 \\ 
 \frac{ 3 }{ 2 }  &  \frac{ 3 }{ 5 }  & 1 \\ 
\end{array}
\right) 
\left( 
\begin{array}{rrr} 
8 &  - 10 &  - 6 \\ 
 - 10 & 50 &  - 15 \\ 
 - 6 &  - 15 & 18 \\ 
\end{array}
\right) 
\left( 
\begin{array}{rrr} 
1 &  \frac{ 5 }{ 4 }  &  \frac{ 3 }{ 2 }  \\ 
0 & 1 &  \frac{ 3 }{ 5 }  \\ 
0 & 0 & 1 \\ 
\end{array}
\right) 
 = \left( 
\begin{array}{rrr} 
8 & 0 & 0 \\ 
0 &  \frac{ 75 }{ 2 }  & 0 \\ 
0 & 0 & 0 \\ 
\end{array}
\right) 
$$
$$ Q^T D Q = H  $$
$$\left( 
\begin{array}{rrr} 
1 & 0 & 0 \\ 
 -  \frac{ 5 }{ 4 }  & 1 & 0 \\ 
 -  \frac{ 3 }{ 4 }  &  -  \frac{ 3 }{ 5 }  & 1 \\ 
\end{array}
\right) 
\left( 
\begin{array}{rrr} 
8 & 0 & 0 \\ 
0 &  \frac{ 75 }{ 2 }  & 0 \\ 
0 & 0 & 0 \\ 
\end{array}
\right) 
\left( 
\begin{array}{rrr} 
1 &  -  \frac{ 5 }{ 4 }  &  -  \frac{ 3 }{ 4 }  \\ 
0 & 1 &  -  \frac{ 3 }{ 5 }  \\ 
0 & 0 & 1 \\ 
\end{array}
\right) 
 = \left( 
\begin{array}{rrr} 
8 &  - 10 &  - 6 \\ 
 - 10 & 50 &  - 15 \\ 
 - 6 &  - 15 & 18 \\ 
\end{array}
\right) 
$$
Algorithm discussed at http://math.stackexchange.com/questions/1388421/reference-for-linear-algebra-books-that-teach-reverse-hermite-method-for-symmetr 
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$$  4 \left(x - \frac{5}{4} y  - \frac{3}{4} z \right)^2 + \frac{75}{4} \left( y  - \frac{3}{5} z\right)^2 = 4x^2 + 25 y^2 + 9 z^2 - 15 yz - 6 zx - 10 xy   $$
$$  \frac{1}{4} \left(4x - 5 y  - 3 z \right)^2 + \frac{3}{4} \left( 5y  - 3 z\right)^2 = 4x^2 + 25 y^2 + 9 z^2 - 15 yz - 6 zx - 10 xy   $$
A: Completing the squares, the quadric equation can be rewritten as $$\frac14\left(4x-5y-3z\right)^2+\frac34(5y-3z)^2=0,$$ so its spectrum is $(+,+,0,0)$, which makes it a collapsed elliptic cylinder, i.e., a line. Examining the above equation, we can see that this line is the intersection of the planes $4x-5y-3z=0$ and $5y-3z=0$, and adding the equation of the other given plane results in a straightforward system of linear equations to solve.
A: Continuing your second method. 
Denote: $a=2x,b=5y,c=3z$, then:
$$\begin{cases}4x^2+25y^2+9z^2-10xy-15yz-6zx=0\\
x+y+z=5\end{cases} \Rightarrow \\
\begin{cases}a^2+b^2+c^2-ab-bc-ca=0\\
\frac a2+\frac b5+\frac c3=5\end{cases} \Rightarrow \\
2a^2+2b^2+2c^2-2ab-2bc-2ca=0 \Rightarrow \\
(a-b)^2+(b-c)^2+(c-a)^2=0 \Rightarrow \\
a=b=c=\frac{150}{31} \Rightarrow \\
(x,y,z)=\left(\frac{75}{31},\frac{30}{31},\frac{50}{31}\right)$$
WA answer.
