(x^n -27) is divisible by: (x-3), when n =?
OPTIONS: ANS:1,2,3,4

• answer will be 3...
  go with the options... (x^1 -27) not divisible by (x-3)
  when we put the the value of n as 3, (x^3 - 27)/(x-3)
  (x^3 - 3^3)/(x-3) as we know the formula of (a^3 -b^3) is (a-b)(a^2+ab+b^2) same way put the formula (x-3)(x^2+3x+3^2)/(x-3) so 3 will be the answer
• 7 years agoHelpfull: Yes(22) No(0)
• ans 3.
  because x to the power 3 and 3 to the power 3 means 27..
• 7 years agoHelpfull: Yes(3) No(2)
• option:3
  (calculate the power as three)
• 7 years agoHelpfull: Yes(1) No(0)
• n=3
  -By using identity formula of cubes i.e. a^3-b^3=(a-b)*(a^2+a*b+b^2)
  therefore, (x^3-27)/(x-3) can be expanded as
  =(x^3-3^3)/(x-3)
  =((x-3)*(x^2+3*x+3^2))/(x-3)
• 7 years agoHelpfull: Yes(1) No(0)
• (x^3-3^3)%(x-3)=0
• 6 years agoHelpfull: Yes(1) No(0)