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?
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
$$\begin{cases} (a-c) = (x-y)k\\ (b-a) = (y-z)k\\ (c-b) = (z-x)k\\ \end{cases}$$
R=QQ[a,b,c]I=ideal(a^2+a*b+b^2-9,b^2+b*c+c^2-52,c^2+c*a+a^2-49)(49*b^2+39*b*c+9*c^2-52*a^2)%I -- not 0– Jan-Magnus Økland May 16 '17 at 16:26