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?

Given:

p = \(\frac{a^2}{(b-a)(c-a)}\), q = \(\frac{b^2}{(c-b)(a-b)}\), r = \(\frac{c^2}{(a-c)(b-c)}\)

Calculation:

Let a = 1, b = 2, c = 3

According to the question,

⇒ p = \(\frac{1^2}{(2-1)(3-1)}\) = \(\frac{1}{2}\)

⇒ q = \(\frac{2^2}{(3-2)(1-2)}\) = -4

⇒ r = \(\frac{3^2}{(1-3)(2-3)}\) = \(\frac{9}{2}\)

Now, (p + q + r)² = (\(\frac{1}{2}\) - 4 + \(\frac{9}{2}\))² = (1)² = 1

∴ The value of (p + q + r)2 is 1.