We are given
$ab+bc+ca = abc$
Hence $bc(a-1) = abc - bc = ab + ac = a(b+c)$
Hence
$\frac{(b+c)}{bc(a-1)} = \frac{(b+c)}{a(b+c)} = \frac{1}{a}$
or
$\frac{(b+c)}{bc(a-1)} = \frac{1}{a}\cdots(1)$
similarly
$\frac{c+a)}{ca(b-1)} = \frac{1}{b}\cdots(2)$
and
$\frac{(a+b)}{ab(c-1)} = \frac{1}{c}\cdots(3)$
Adding (1),(2),(3) we get
$\frac{(b+c)}{bc(a-1)} + \frac{(c+a)}{ca(b-1)} + \frac{(a+b)}{ab(c-1)}$
$=\frac{1}{a} + \frac{1}{b} + \frac{1}{c}$
$=\frac{bc+ca+ab}{abc} = \frac{abc}{abc}=1$
We are given $x^3+3367=2^n$
We know $3367=7 * 13 &*37
So let is work $x^3=2^n \pmod 7$
Working in mod 7 we have $x^3 \in \{1,-1,0\}$ and $2^n \in \{1,2,4\}$ so we get 1 as common and for that n has to be multiple of 3.
So we get
$x^3= 2^{3k} \pmod 7$
Going back to the original equation we get
$x^3 + 3367 = 2^{3k}$
Or $(2^k)^3 - x^3 = 3367$
Or $y^3 -x^3 = 3367$ where $y = 2^k$
As $(y-x) | y^3-x^3$ so $y-x | 3367\cdots(1)$
Further
As we know $(y-x)^3 = y^3-3y^2x + 3yx^2 - x^3 = (y^3-x^3) - 3yx(y-x) \lt y^3-x^3$ when $y-x \gt 0$
So $y-x\lt 15\cdots(2)$
So $y-x \in \{1,7\}\cdots(2)$ from (1) and (2)
$y-x =1$
gives $x^3 + 3367 = (x+1)^3 = x^3 + 3x^2 + 3x + 1$
Or $3367 = 3x^2+ 3x + 1$
Or $3x^2+3x = 3366$
Or $x^2 + x = 1122$
x = 33 and y = 34 and y is not power of 2 so not a solution
$y-x=7$
gives
$x^3 + 3367 = x^3 + 21x^2 + 147 x + 343$
or $3024 = 21x^2+ 147x
or $x^2+ 7x = 144 $
or $x = 9$ or $x=-16$
only positive value admissible giving x = 9 and y = 16
$2^n = y^3$ giving $2^n = 2^{12}$ or n = 12
Hence x=3 , n = 12
