If (x -1)2 + (y – 2)4 + (z – 3)6 = 0 then the value of xyz is …�

1.	If (x -1)2 + (y – 2)4 + (z – 3)6 = 0 then the value of xyz is ………………..A) 1 B) 6 C) 14 D) 0
Answer» Correct option is (B) 6 \((x-1)^2+(y-2)^4+(z-3)^6 = 0\) \(\because\) Square of any real number is always greater or equal to zero. i.e., \(a^2\geq0\) and sum of positive numbers is always positive. \(\therefore\) \((x-1)^2+(y-2)^4+(z-3)^6 = 0\) is only possible when x-1 = 0, y-2 = 0 and z-3 = 0 \(\Rightarrow\) x = 1, y = 2 and z = 3 \(\therefore\) Value of xyz \(=1\times2\times3=6\) Correct option is B) 6

If (x -1)2 + (y – 2)4 + (z – 3)6 = 0 then the value of xyz is ………………..A) 1 B) 6 C) 14 D) 0

Answer»

Correct option is (B) 6

\((x-1)^2+(y-2)^4+(z-3)^6 = 0\)

\(\because\) Square of any real number is always greater or equal to zero.

i.e., \(a^2\geq0\) and sum of positive numbers is always positive.

\(\therefore\) \((x-1)^2+(y-2)^4+(z-3)^6 = 0\) is only possible when

x-1 = 0, y-2 = 0 and z-3 = 0

\(\Rightarrow\) x = 1, y = 2 and z = 3

\(\therefore\) Value of xyz \(=1\times2\times3=6\)

Correct option is B) 6