(x-y)^n-(y-z)^n+z^n always divisible by what? given x=yz.
options:
1)yz
nt rember other options.

• here its given: x=yz, so the equation simplifies to :
  (yz-y)^n-(y-z)^n+z^n
  
  now substitute n=1:
  then we get (yz-y)-(y-z)+z == yz
  
  now substitute n=2:
  then we get (yz-y)^2-(y-z)^2+z^2 == (yz)^2+zy^2+2yz
  so this is also divisible by yz
  
  try 3,4 and so on....
  the above equation would be divisible from yz.
  
  hence ans:1)yz
• 9 years agoHelpfull: Yes(9) No(16)
• option
  a)yz
  b)z but not y
  
  answer is b) z but not y
• 9 years agoHelpfull: Yes(5) No(6)
• @ rakesh can u solve this???
  
• 9 years agoHelpfull: Yes(3) No(3)
• x=yz so(yz-y)^n-(y-z)^n+z^n={(yz)^n-......+(y)^n}-{(y)^n-.......+(z)^n}+(z)^n
  =(yz)^n-.......+(y)^n-(y)^n+.........-(z)^n+(z)^n
  If we solve this equation we see the result is always divisible by yz...
  so the answer is option 1)yz
• 9 years agoHelpfull: Yes(3) No(4)
• b)z but not y
• 9 years agoHelpfull: Yes(1) No(0)
• as x=yz
  ((x-y)^n-(y-z)^n+z^n)/???
  givn x=yz
  nw ((yz-y)^n-(y-z)^n+z^n)/??
  z^n((y-y/z)^n-(y/z-1)^n+1)/??
  i think divisible by Z nt y
• 9 years agoHelpfull: Yes(1) No(1)
• [x^n(+-)y^n-(y^n(+-)z^n)+z^n]-(.......)within this bracket whatever terms are there all contain both y and z,so this part divisible by y or z or yz.now the part within square bracket depend upon whether the value of n is even or odd(+ or _ depend upon this),if n is even then 1st part become
  x^n+y^n-y^n-z^n+z^n=x^n=(yz)^n thus divisible by yor z or yz.
  
  now,if n is odd,
  x^n-y^n-y^n+z^n+z^n.now it is divisible by what?
  
  upon this 1st part only ans is depended,because 2nd part always divisible by y or z or yz.
• 9 years agoHelpfull: Yes(0) No(6)
• 1,as put x=yz in equ^n we get it is always divisible by yz
• 9 years agoHelpfull: Yes(0) No(2)
• if n is odd then (yz^n............-y^n)-(y^n.........-z^n)+z^n
  yz^n.....-2y^n......+2z^n then how will it be divided by yz??
  can any bodu explain.........
• 9 years agoHelpfull: Yes(0) No(0)
• take binomial expansion of (x-y)^n-(y-z)^n+z^n
  then you will see divisibility depends on the nature of "n" .
  ie . if n-even
  then the whole expression will be divisible y;
  and if n-odd
  then there you can not reach to the result whether it gets divisible or not.
  
  so in my opinion question is "aadha-adhura"
• 9 years agoHelpfull: Yes(0) No(1)