The value of $$\frac{{{{\left( {x - y} \right)}^3} + {{\left( {y - z} \right)}^3} + {{\left( {z - x} \right)}^3}}}{{{{\left( {{x^2} - {y^2}} \right)}^3} + {{\left( {{y^2} - {z^2}} \right)}^3} + {{\left( {{z^2} - {x^2}} \right)}^3}}}$$ is = ?
A. 0
B. 1
C. $${\left[ {2\left( {x + y + z} \right)} \right]^{ - 1}}$$
D. $${\left[ {\left( {x + y} \right)\left( {y + z} \right)\left( {z + x} \right)} \right]^{ - 1}}$$
Answer: Option D
Solution(By Examveda Team)
$$\eqalign{ & {\text{Since }}\left( {x - y} \right) + \left( {y - z} \right) + \left( {z - x} \right) = 0 \cr & {\text{So,}}{\left( {x - y} \right)^3} + {\left( {y - z} \right)^3} + {\left( {z - x} \right)^3} \cr & = 3\left( {x - y} \right)\left( {y - z} \right)\left( {z - x} \right) \cr & {\text{Since}}\left( {{x^2} - {y^2}} \right) + \left( {{y^2} - {z^2}} \right) + \left( {{z^2} - {x^2}} \right) = 0 \cr & {\text{So,}}\left( {{x^2} - {y^2}} \right) + \left( {{y^2} - {z^2}} \right) + \left( {{z^2} - {x^2}} \right) \cr & = 3\left( {{x^2} - {y^2}} \right)\left( {{y^2} - {z^2}} \right)\left( {{z^2} - {x^2}} \right) \cr & \therefore {\text{Given expression }} \cr & {\text{ = }}\frac{{3\left( {x - y} \right)\left( {y - z} \right)\left( {z - x} \right)}}{{3\left( {{x^2} - {y^2}} \right)\left( {{y^2} - {z^2}} \right)\left( {{z^2} - {x^2}} \right)}} \cr & = \frac{1}{{\left( {x + y} \right)\left( {y + z} \right)\left( {z + x} \right)}} \cr & = {\left[ {\left( {x + y} \right)\left( {y + z} \right)\left( {z + x} \right)} \right]^{ - 1}} \cr} $$Related Questions on Simplification
A. 20
B. 80
C. 100
D. 200
E. None of these
A. Rs. 3500
B. Rs. 3750
C. Rs. 3840
D. Rs. 3900
E. None of these
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