If m^2-8m+1=0,then m^3+m^(-3)= ?

• here from equation 1 we will get
  m^2-8m+1=0
  m^2=8m-1
  m=(8m-1)/m.........(1)
  m=8-1/m
  so m+1/m=8
  now in equation 2
  m^3+1/m^3
  (m+1/m)(m^2 - m*1/m + 1/m^2)
  (m+1/m)(m^2 - 1 + 1/m^2)
  (m+1/m)(m^2 + 1/m^2 - 1)
  (m+1/m){(m+1/m)^2 - 2m*1/m - 1} {since a^2+b^2=(a+b)^2-2ab}
  (m+1/m){(m+1/m)^2 - 2-1}
  now putting the value of m from equation 1
  8(8^2-2-1)
  8(64-3)
  =488
• 9 years agoHelpfull: Yes(23) No(0)
• m^2-8m+1=0
  Now , m^2=8m -1
  Dividing both sides by m,we get
  m =8 - (1/m)
  So,
  m^3 +1/m^3
  =(m+1/m)(m^2 - 1 + 1/m^2) {bcoz a^3 +b^3= (a+b)(a^2 - ab + b^2)}
  =8{(m+1/m)^2 - 4} As m+1/m =8 and [a2+b2 - 2ab=(a-b)2=(a+b)2-4ab)]
  =8[8^2-4]
  =8[64-4]
  =480 Ans
  
• 9 years agoHelpfull: Yes(4) No(2)
• ans:-488
  m^3+m^-3
  (m^6+1)/m^3
  (m^2+1)(m^4+1-m^2)/m^3
  (8m)(62m^2-m^2)/m^3 (m^2+1=8m from if condition and square both side then m^4+1=62 m^2)
  8*61
  488
• 9 years agoHelpfull: Yes(3) No(0)
• m^2-8m+1=0
  m say div
  m+1/m=8 -1
  m^3+m^(-3)= (m+1/m)^3 -3(m+1/m)
  =(8)^3 - 3*8
  =488
• 9 years agoHelpfull: Yes(1) No(0)
• m2-8m+1=0
  first find value of m=(4+root(15),4-root(15))
  m3+1/m3=(m+m)(m2-m2+m2)
  = 2m(m2 )
  =2m3
  =2[4+root(15)]^3
  After solving we get value of
  m3+1/m3= 128+30root(15) Ans
  
  
• 9 years agoHelpfull: Yes(0) No(2)
• Ans : 488
  
  
  
  
• 9 years agoHelpfull: Yes(0) No(1)