If x = a + m, y = b + m, z = c + m, then the value of $$\frac{{{x^2} + {y^2} + {z^2} - yz - zx - xy}}{{{a^2} + {b^2} + {c^2} - ab - bc - ca}}$$ is = ?
A. 1
B. $$\frac{{x + y + z}}{{a + b + c}}$$
C. $$\frac{{a + b + c}}{{x + y + z}}$$
D. Not possible to find
Answer: Option A
Solution(By Examveda Team)
$$\eqalign{ & \frac{{{x^2} + {y^2} + {z^2} - yz - zx - xy}}{{{a^2} + {b^2} + {c^2} - ab - bc - ca}}{\text{ }} \cr & = \frac{{{{\left( {a + m} \right)}^2} + {{\left( {b + m} \right)}^2} + {{\left( {c + m} \right)}^2} - \left( {b + m} \right)\left( {c + m} \right) - \left( {c + m} \right)\left( {a + m} \right) - \left( {a + m} \right)\left( {b + m} \right)}}{{{a^2} + {b^2} + {c^2} - ab - bc - ca}} \cr & = \frac{{{a^2} + {m^2} + 2am + {b^2} + {m^2} + 2bm + {c^2} + {m^2} + 2cm - bc - bm - cm - {m^2} - ca - cm - am - {m^2} - ab - am - bm - {m^2}}}{{{a^2} + {b^2} + {c^2} - ab - bc - ca}} \cr & = \frac{{{a^2} + {b^2} + {c^2} - ab - bc - ca}}{{{a^2} + {b^2} + {c^2} - ab - bc - ca}} \cr & = 1 \cr} $$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
Join The Discussion