<s>[INST] find the value $\frac{49b^2+39bc+9c^2}{a^2}=52$ Let $a,b.c$ be real numbers such that
$$\begin{cases}a^2+ab+b^2=9\\
b^2+bc+c^2=52\\
c^2+ca+a^2=49
\end{cases}$$
show that
$$\dfrac{49b^2+39bc+9c^2}{a^2}=52$$
I have found this problem solution by geomtry methods.solution 1,can you someone have Algebra methods?
 [/INST] Hint:
$$\begin{cases}a^2+ab+b^2=x\\
b^2+bc+c^2=y\\
c^2+ca+a^2=z
\end{cases}$$
Then 
$$\begin{cases}
(a-c)(a+b+c) = x-y\\
(b-a)(a+b+c) = y-z\\
(c-b)(a+b+c) = z-x\\
\end{cases}$$
Let $(a+b+c)^{-1}=k$, then 
$$\begin{cases}
(a-c) = (x-y)k\\
(b-a) = (y-z)k\\
(c-b) = (z-x)k\\
\end{cases}$$
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