<s>[INST] Find the value of $ ( ab + bc + ca )^2 $ If $a,b,c$ are real numbers which satisfy 
$a^2+b^2+ab  = 9$
$b^2+c^2+bc  = 16$
$c^2+a^2+ca  = 25$
find the value of $ ( ab + bc + ca )^2 $
 [/INST] let 
since
$$\begin{cases}a^2-2a\cdot b\cos{\dfrac{2\pi}{3}}+b^2=9\\
b^2-2bc\cos{\dfrac{2\pi}{3}}+c^2=16\\
c^2-2ca\cos{\dfrac{2\pi}{3}}+a^2=25
\end{cases}
$$
then we let
$$PA=a,PB=b,PC=c,\angle APB=\angle BPC=\angle APC=\dfrac{2\pi}{3}$$
By cosine 
$$|AB|^2=25,|BC|^2=16,|AC|^2=9$$
so $\angle C=\dfrac{\pi}{2}$,
then we have
$$S_{ABC}=S_{APC}+S_{BPC}+S_{APB}=\dfrac{1}{2}\sin{\dfrac{2\pi}{3}}(ab+bc+ac)$$
$$\Longrightarrow \dfrac{\sqrt{3}}{4}(ab+bc+ac)=\dfrac{1}{2}|AC||BC|=6$$
so
$$ab+bc+ac=(8\sqrt{3})^2=192$$
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