Answer & Solution

Answer: Option B

Solution:

$$\frac{{{{\left( {a - b} \right)}^2}}}{{\left( {b - c} \right)\left( {c - a} \right)}} + $$ $$\frac{{{{\left( {b - c} \right)}^2}}}{{\left( {a - b} \right)\left( {c - a} \right)}} + $$ $$\frac{{{{\left( {c - a} \right)}^2}}}{{\left( {a - b} \right)\left( {b - c} \right)}}$$
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 1^st term
Multiply divide by (b - c) in 2^nd term
Multiply divide by (c - a) in 3^rd term
$$ \Rightarrow \frac{{{{\left( {a - b} \right)}^2}\left( {a - b} \right)}}{{\left( {b - c} \right)\left( {c - a} \right)\left( {a - b} \right)}} + $$ $$\frac{{{{\left( {b - c} \right)}^2}\left( {b - c} \right)}}{{\left( {a - b} \right)\left( {b - c} \right)\left( {c - a} \right)}} + $$ $$\frac{{{{\left( {c - a} \right)}^2}\left( {c - a} \right)}}{{\left( {a - b} \right)\left( {b - c} \right)\left( {c - a} \right)}}$$
$$\eqalign{ & {\text{Let, }}a - b = x \cr & b - c = y \cr & c - a = z \cr & \therefore x + y + z = 0 \cr & \therefore {x^3} + {y^3} + {z^3} = 3xyz \cr & \therefore {\left( {a - b} \right)^2} + {\left( {b - c} \right)^2} + {\left( {c - a} \right)^2} \cr & = 3\left( {a - b} \right)\left( {b - c} \right)\left( {c - a} \right) \cr & \therefore \frac{{3\left( {a - b} \right)\left( {b - c} \right)\left( {c - a} \right)}}{{\left( {a - b} \right)\left( {b - c} \right)\left( {c - a} \right)}} \cr & = 3 \cr} $$

The value of the expression ( a - b )^2 ( b - c ) ( c - a ) + ( b - c )^2 ( a - b ) ( c - a ) + ( c - a )^2 ( a - b ) ( b - c ) = ?

Solution