I might be wrong, but it seems like the answer imbedded in the question is wrong based on the specified assumptions
a^2 + b^2 + c^2 - ab -bc - ca
First we can combine a^2 and b^2 together using the square of difference (a-b)^2 = a^2 -2ab +b^2
Thus a^2 + b^2 - ab part of the equation can be expressed as (a-b)^2 + ab (the plus ab term is needed to make -2ab equal to -ab)
Then we are left with: (a-b)^2 + ab -bc - ca + c^2
This can be rewritten as: (a-b)^2 + b(a -c) + c(c-a)
Rewrite c(c-a) as -c(a-c) ; yielding (a-b)^2 + b(a -c) - c(a-c)
Since the second and third term of the equation contain a common element (a-c) we can further rewrite as:
(a-b)^2 + (b-c)(a -c)
Now, the (a-b)^2 is always positive regardless of values of a or b (because squared)
(b-c) and (a-c) terms can also be positive when a and b are larger than c
However, nowhere in the problem is it indicated that a, b, and c are distinct numbers, or that they cannot equal to each other.
Thus one of the solutions to the equation is the scenario where a=b=c
Under these conditions the entire polynomial will equal to zero
Thus without additional specifications, the answer should be D (not enough info provided)
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