The value of the expression $$\frac{{{{\left( {a - b} \right)}^2}}}{{\left( {b - c} \right)\left( {c - a} \right)}} + $$ $$\frac{{{{\left( {b - c} \right)}^2}}}{{\left( {a - b} \right)\left( {c - a} \right)}} + $$ $$\frac{{{{\left( {c - a} \right)}^2}}}{{\left( {a - b} \right)\left( {b - c} \right)}}$$ = ?
A. 0
B. 3
C. $$\frac{1}{3}$$
D. 2
Answer: Option B
Solution(By Examveda Team)
$$\frac{{{{\left( {a - b} \right)}^2}}}{{\left( {b - c} \right)\left( {c - a} \right)}} + $$ $$\frac{{{{\left( {b - c} \right)}^2}}}{{\left( {a - b} \right)\left( {c - a} \right)}} + $$ $$\frac{{{{\left( {c - a} \right)}^2}}}{{\left( {a - b} \right)\left( {b - c} \right)}}$$Now,
$$ \Rightarrow \frac{{{{\left( {a - b} \right)}^2}}}{{\left( {b - c} \right)\left( {c - a} \right)}} \times \frac{{\left( {a - b} \right)}}{{a - b}}$$
Multiply divide by (a - b) in 1st term
Multiply divide by (b - c) in 2nd term
Multiply divide by (c - a) in 3rd term
$$ \Rightarrow \frac{{{{\left( {a - b} \right)}^2}\left( {a - b} \right)}}{{\left( {b - c} \right)\left( {c - a} \right)\left( {a - b} \right)}} + $$ $$\frac{{{{\left( {b - c} \right)}^2}\left( {b - c} \right)}}{{\left( {a - b} \right)\left( {b - c} \right)\left( {c - a} \right)}} + $$ $$\frac{{{{\left( {c - a} \right)}^2}\left( {c - a} \right)}}{{\left( {a - b} \right)\left( {b - c} \right)\left( {c - a} \right)}}$$
$$\eqalign{ & {\text{Let, }}a - b = x \cr & b - c = y \cr & c - a = z \cr & \therefore x + y + z = 0 \cr & \therefore {x^3} + {y^3} + {z^3} = 3xyz \cr & \therefore {\left( {a - b} \right)^2} + {\left( {b - c} \right)^2} + {\left( {c - a} \right)^2} \cr & = 3\left( {a - b} \right)\left( {b - c} \right)\left( {c - a} \right) \cr & \therefore \frac{{3\left( {a - b} \right)\left( {b - c} \right)\left( {c - a} \right)}}{{\left( {a - b} \right)\left( {b - c} \right)\left( {c - a} \right)}} \cr & = 3 \cr} $$
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It is wrong
P
It's not (a-b)² +.....².......²
it's (a-b)³+......³........³