[(b–c)/a + (c–a)/b + (a–b)/c].[a/(b–c) + b/(c–a) + c/(a–b)]

Some hints for this puzzle are provided  in terms of  the comments furnished below.<o:p></o:p>

1. At the outset, we observe that; a ,b and c cannot be all equal for obvious reasons ( tip: observe the denominators). Each of a, b and c must be non-zero. ( for same reason). Since, the problem asks for a definite solution, it follows that a,b and c cannot be all equal and each of  a,b and c must be non-zero. <o:p></o:p>
2. Let us denote the given expression as S*T; where S = (b – c)/a + (c – a)/b + (a – b)/c; and T = a/(b – c) + b/(c – a) + c/(a – b)<o:p></o:p>
3. Expand S and factorize its numerator, the denominator being a*b*c.<o:p></o:p>
4.  Substitute p = a-b; q = b-c and r = c-a. Observe how each of a, b ,c  comes out in terms of p, q and r in view of  a+b+c = 0. Now, expand T in terms of p, q and r. Factorize numerator of  T ( in terms of p, q and r).<o:p></o:p>
5. Express T in terms of a,b and c.<o:p></o:p>
6. Finally evaluate S*T  to  obtain the desired result, which is a one-digit positive integer.<o:p></o:p>

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