Question : The value of the expression $\frac{(a-b)^{2}}{(b-c)(c-a)}+\frac{(b-c)^{2}}{(a-b)(c-a)}+\frac{(c-a)^{2}}{(a-b)(b-c)}$ is:
Option 1: $0$
Option 2: $3$
Option 3: $\frac{1}{3}$
Option 4: $2$
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Correct Answer: $3$
Solution :
Given: $\frac{(a-b)^{2}}{(b-c)(c-a)}+\frac{(b-c)^{2}}{(a-b)(c-a)}+\frac{(c-a)^{2}}{(a-b)(b-c)}$
Multiplying the numerator and denominator of each fraction by $(a-b)$, $(b-c)$ and $(c-a)$ respectively:
$\frac{(a-b)^{2}×(a-b)}{(b-c)(c-a)(a-b)}+\frac{(b-c)^{2}(b-c)}{(a-b)(c-a)(b-c)}+\frac{(c-a)^{2}(c-a)}{(a-b)(b-c)(c-a)}$
= $\frac{(a-b)^{3}+(b-c)^{3}+(c-a)^{3}}{(a-b)(b-c)(c-a)}$
We know that $A^3+B^3+C^3=3ABC$ when $A + B + C = 0$
= ${\frac{3(a-b)(b-c)(c-a)}{(a-b)(b-c)(c-a)}}$
= $3$
Hence, the correct answer is $3$.
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