<s>[INST] Finding value of $\frac{49b^2-33bc+9c^2}{a^2}$ 
If $a,b,c$ are positive real numbers such that
$a^2+ab+b^2=9, b^2+bc+c^2=52,$
$c^2+ac+a^2=49$. Then finding $\displaystyle \frac{49b^2-33bc+9c^2}{a^2}$ is

Try: let $O$ be a point inside the Triangle $ABC$ such that angle $OAB$ and angle $OBC$ and $OCA$ is $120^\circ$. So we have
$OA=a,OB=b,OC=c$ and using cosine formula we have $AB=3,BC=7,CA=\sqrt{52}$.
Now using Area of triangle is $\displaystyle ab+bc+ca=\frac{42}{\sqrt{3}}$.
Could some help me to solve it , Thanks
 [/INST] Applying cosine rule on  $\triangle OBC$, we have
$$BC=\sqrt{b^2+bc+c^2}=\sqrt{52}$$
Similarly,
$$CA=7$$
By applying cosine rule on  $\triangle ABC$
$$BC^2=AB^2+AC^2-2AB\cdot AC \cos \angle CAB$$
$$52=9+49-2(3)(7)\cos \angle CAB$$
$$\cos \angle CAB =\frac{1}{7} $$
Let's apply inversion at point $A$ with radius $1$, and let $O'$ be the image of $O$, $B'$ be the image of $B$ and $C'$ be the image of $C$.
Then $\triangle ABC \sim \triangle AC'B'$ , $\triangle AOB \sim \triangle AB'O'$,  and $\triangle AOC \sim \triangle AC'O'$.
Hence
$$O'B'=\frac{BO \cdot AB' }{AO}=\frac{b\cdot AB'\cdot AB}{AO\cdot AB}=\frac{b}{3a}$$
$$O'C'=\frac{CO \cdot AC' }{AO}=\frac{c\cdot AC'\cdot AC}{AO\cdot AC}=\frac{c}{7a}$$
$$B'C'=\frac{CB \cdot AC' }{AB}=\frac{CB\cdot AC'\cdot AC}{AC\cdot AB}=\frac{CB}{7\cdot 3}=\frac{\sqrt{52}}{21}$$
Let's compute $\cos \angle B'O'C'$, from quadrilateral $AO'B'C'$.
$$\angle AB'O'+ \angle B'O'C'+\angle O'C'A + \angle C'AB'= 2\pi $$
$$\angle AOB+ \angle B'O'C'+\angle COA + \angle CAB= 2\pi $$
$$\frac{2\pi}3+ \angle B'O'C'+ \frac{2\pi}3+ \angle CAB= 2\pi $$
$$ \angle B'O'C'= \frac{2\pi}3 - \angle CAB$$
\begin{align}\cos \angle B'O'C' &= -\frac12 \cos \angle CAB +\frac{\sqrt3}2 \sin \angle CAB\\
&=-\frac1{14} +\frac{\sqrt3}2 \frac{\sqrt{48}}{7}\\
&= \frac{11}{14}\end{align}
Applying cosine rule on $\triangle O'B'C'$,
$$B'C'^2=O'B'^2+O'C'^2-2\cdot O'B'\cdot O'C' \cos \angle B'O'C'$$
$$\frac{52}{21^2}=\left(\frac{b}{3a} \right)^2+\left(\frac{c}{7a} \right)^2-2\cdot \left(\frac{b}{3a} \right)\cdot \left(\frac{c}{7a} \right) \cdot \frac{11}{14}$$
$$\frac{52}{7^2}=\left(\frac{b}{a} \right)^2+\left(\frac{3c}{7a} \right)^2-2\cdot \left(\frac{b}{a} \right)\cdot \left(\frac{c}{7a} \right) \cdot \frac{33}{14}$$
$$\frac{52}{7^2}=\left(\frac{7b}{7a} \right)^2+\left(\frac{3c}{7a} \right)^2-\frac{33bc}{(7a)^2}$$
$$\frac{(7b)^2+(3c)^2-33bc}{a^2}=52$$
Remark: 
In general, if $$a^2+ab+b^2=\gamma$$
$$b^2+bc+c^2=\alpha$$
$$c^2+ac+a^2=\beta$$
then 
$$a^2\alpha =\beta b^2 + \gamma c^2 - Dbc $$
where $$D=\left[ - \frac{ \beta + \gamma - \alpha}{2} \right] + \frac{\sqrt3}{2} \sqrt{4 \beta \gamma - (\alpha+\beta-\gamma)^2}$$
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