If a cubic number y is added to a perfect square x and resultant sum is divided by a cubic number z we get result equal to y. Find x,y & z considering they are distinct and numbers are within 1 to 100.

• Pls check if given statement is correct.
  If a cubic number y is added to a perfect square x and resultant sum is divided by a cubic number z we get result equal to y. Find x,y & z considering they are distinct and numbers are within 1 to 100.
  
  If the statement is read as
  If a cubic number y is added to a perfect square x and resultant sum is divided by a 'perfect square number' z we get result equal to y. Find x,y & z considering they are distinct and numbers are within 1 to 100.
  
  then (y+x)/z=y
  (2^3+ 8^2)/(3^2) = (8+64)/9= 72/9 =8 = 2^3 =y
• 11 years agoHelpfull: Yes(3) No(1)
• By considering 0 is a perfect square, answer will be
  
  So many possibles
  
  x=0; y=8(cube of 2); z=1(cube of 1)--> (8+0)/1 = 8
  x=0; y=27(cube of 3); z=1(cube of 1)--> (27+0)/1 = 27
  x=0; y=64(cube of 4); z=1(cube of 1)--> (64+0)/1 = 64
  
  Other than these, there is no possible i think.
  
• 11 years agoHelpfull: Yes(1) No(3)
• Question may wrong or typical mistake.
  
  Otherwise,
  
  If 0 is a perfect square, answer will be
  
  Three possibles
  x=0;y=8;z=1
  x=0;y=27;z=1
  x=0;y=64;z=1
  
  i.e,
  x=0; y=8(cube of 2); z=1(cube of 1)--> (8+0)/1 = 8
  x=0; y=27(cube of 3); z=1(cube of 1)--> (27+0)/1 = 27
  x=0; y=64(cube of 4); z=1(cube of 1)--> (64+0)/1 = 64
  
  Other than these, there is no possible i think.
• 11 years agoHelpfull: Yes(1) No(2)