<s>[INST] Find the value of $ab+ 2cb+\sqrt3 ac$? Three positive real numbers $a,b,c$ satisfy the equations $a^2+\sqrt3 ab+b^2=25$,  $b^2+c^2=9$  and $a^2+ac+c^2=16$ .Then find the value of $ab+ 2cb+\sqrt3 ac$?
Is there some way to find the desired value without actually finding values of $a,b,c$ or any other smart method to find $a,b,c$        
 [/INST] The given system is $$a^2+\sqrt 3ab+b^2=25\\b^2+c^2=9\\a^2+ac+c^2=16$$
Then $$9+16=25\iff a^2+ac+c^2+ b^2+c^2= a^2+\sqrt 3ab+b^2\iff c(a+2c)=\sqrt 3 ab$$
It follows $$ab+2cb+\sqrt 3 ac=b(a+2c)+\sqrt 3ac=\frac{\sqrt 3\space a(b^2+c^2)}{c}=\frac {9\sqrt 3\space a}{c}=X$$
Hence one needs to calculate $\frac ac$. The successive following steps are clear to understand:
$$a^2+\sqrt 3\space ab+(9-c^2)=25\iff  b=\frac {16+c^2-a^2}{\sqrt 3\space a}=\frac{ac+2c^2}{\sqrt 3\space a}$$ It follows $$ \left(\frac{ac+2c^2}{\sqrt 3\space a}\right)^2+c^2=9$$ 
So the system $$4c^4+4ac^3+4a^2c^2=27a^2\\a^2+ac+c^2=16$$
Hence $$4c^2(16)=27a^2\Rightarrow \frac ac=\frac{8}{3\sqrt 3}$$ Consequently $$X=\frac{9\sqrt 3\cdot 8}{3\sqrt 3}=\color{red}{24}$$
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