Question : The numerical value of $\frac{(a–b)^{2}}{(b–c)(c–a)}+\frac{(b–c)^{2}}{(c–a)(a–b)}+\frac{(c–a)^{2}}{(a–b)(b–c)}$ is: $(a\neq b\neq c)$
Option 1: $0$
Option 2: $1$
Option 3: $\frac{1}{3}$
Option 4: $3$
Correct Answer: $3$
Solution :
Given: $\frac{(a–b)^{2}}{(b–c)(c–a)}+\frac{(b–c)^{2}}{(c–a)(a–b)}+\frac{(c–a)^{2}}{(a–b)(b–c)}$
Taking LCM of the given expression, we have,
= $\frac{(a–b)(a–b)^{2}+(b–c)(b–c)^{2}+(c–a)(c–a)^{2}}{(a–b)(b–c)(c–a)}$
= $\frac{(a–b)^{3}+(b–c)^{3}+(c–a)^{3}}{(a–b)(b–c)(c–a)}$
We know that $x^{3}+y^{3}+z^{3}=3xyz$, if $x+y+z=0$
According to this, we have,
⇒ $(a-b)=x$, $(b-c)=y$, $(c-a)=z$
So, $x+y+z=(a-b+b-c+c-a)=0$
Now, $\frac{(a–b)^{3}+(b–c)^{3}+(c–a)^{3}}{(a–b)(b–c)(c–a)}$
= $\frac{3(a–b)(b–c)(c–a)}{(a–b)(b–c)(c–a)}$
= 3
Hence, the correct answer is $3$.
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