<s>[INST] if $6^m+2^{m+n}\cdot 3^w+2^n = 332\;,$ Then $m^2+mn+n^2 = \;,$ where $m,n,w \in \mathbb{Z^{+}}$ Suppose $m.n$ are positive integers such that $6^m+2^{m+n}\cdot 3^w+2^n = 332\;,$ Then $m^2+mn+n^2 = $
$\bf{My\; Try::}$ Given $6^m+2^{m+n}\cdot 3^w+2^n=332\Rightarrow 2^m\cdot 3^n+2^{m}\cdot 3^n\cdot 3^w+2^n=332=2^2\cdot 133$
$\Rightarrow 2^m\cdot \left(3^n+3^{n}\cdot 3^{w}+2^{n-m}\right)=2^2\cdot 133\;,$ Using Camparasion, we get $m=2$
$\Rightarrow (3^n+3^m\cdot 9+2^{n-2}) = 133$
Now How Can i solve after that, Help me
 [/INST] You made some arithmetic errors in what you did (the initial factorization is incorrect, plus $332 = 4\cdot 83$, not $4\cdot 133$) , but you are on the right track.
$$
  6^m + 2^{m+n}\cdot 3^w + 2^n = 332 \Rightarrow
  2^m\cdot 3^m + 2^m\cdot 2^n \cdot 3^w + 2^n = 332.
$$
Now assuming $m\le n$ (if it doesn't work, you must also try $n\le m$), this is equal to
$$2^m(3^m + 2^n\cdot 3^w + 2^{n-m}) = 2^2\cdot 83,$$
so that as you say $m=2$, and then
$$9 + 2^n\cdot 3^w + 2^{n-2} = 83 \Rightarrow 2^{n-2}(4\cdot 3^w + 1) = 74.$$
Thus $n=3$ (and $w=2$).
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