If p=\frac{a^{2}}{(b-a)(c-a)}, q=\frac{b^{2}}{(c-b)(a-b)}, r=\frac{c^{2}}{(a-c)(b-c)} then what is (p+q+r)^{2} equal to?

Explanation

Taking a common denominator (a-b)(b-c)(c-a), we rewrite p, q, r with matched signs. p = \frac{a^2}{(a-b)(a-c)} = \frac{-a^2(b-c)}{(a-b)(b-c)(c-a)}. Similar adjustments for q and r yield the numerator -a^2(b-c)-b^2(c-a)-c^2(a-b). This numerator factorizes to -(a-b)(b-c)(c-a). Thus, p+q+r = \frac{-(a-b)(b-c)(c-a)}{(a-b)(b-c)(c-a)} = -1. Therefore, (p+q+r)^2 = (-1)^2 = 1.

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