(a^4)+(1/a^4)=119
then (a^3)+(1/a^3)=?

Don't remember the options.

• answer is 10*13^1/2
  a^4 + 1/a^4 = (a^2 + 1/a^2)^2 -2(a^2)(1/a^2)
  from above we get a^2 + 1/a^2 = 11
  a^2 + 1/a^2 = (a + 1/a)^2 - 2(a)(1/a)
  from this we get a+1/a= (13)^1/2
  
  a^3 + 1/a^3 = (a +1/a)^3 -3(a)(1/a)[a + 1/a]
  = 13*3^1/2 - 3*13^1/2
  = 10*13^1/2
  
• 11 years agoHelpfull: Yes(9) No(0)
• The answer is 130.
  
  Well there are two ways 2 solve this
  
  The Regular calculation is,
  
  (a^2)^2+(1/a^2)^2 =119 [since a^2+b^2 = (a+b)^2-2ab)]
  (a^2+1/a^2)^2 - 2(a^2)(1/a^2) = 119
  (a^2+1/a^2)^2 = 121
  a^2+1/a^2=11 [since a^2+b^2 = (a+b)^2-2ab)] ----- (1)
  (a+1/a)-2(a)(1/a) =11
  (a+1/a) = 13 --------(2)
  
  a^3+(1/a)^3 = (a+1/a)(a^2+(1/a)^2- a(1/a) )[since a^3+b^3 = (a+b)(a^2+b^2-ab)]
  = (13)(11-1)[From (1) and (2) (a+1/a)=13 & a^2+1/a^2=11 ]
  = (13)(10)
  = 130
  
  
  The short cut for the problem for any number in place of 119 is lets say n,
  add 2 to the number and take square root of that number (sqrt(n+2))
  and the answer is (sqrt(n+2)-1)(sqrt(n+2)+2) is the answer.
• 11 years agoHelpfull: Yes(2) No(4)