(40*40*40 - 31*31*31)/(40*40+40*31+31*31)=?
option
a) 8
b) 9
c) 71
d) 51

• This question is based on this....
  
  (a^3 - b^3)/(a^2 + ab + b^2) where a=40 and b=31
  
  a^3 - b^3 = (a-b)(a^2 + b^2 + ab)
  
  so we get (a-b)(a^2 + b^2 + ab)/(a^2 + b^2 + ab) where a=40 and b=31
  
  now finally we get (a-b) so (40 - 31)= 9
  
  therefore option (b) will be the answer
• 11 years agoHelpfull: Yes(27) No(0)
• let a=40 and b=31 now we get (a^3)-(b^3)/(a^2+a*b+b^2)=a-b.
  so answer is 40-31=9
• 11 years agoHelpfull: Yes(2) No(0)
• Ans:c
  sol: (a3+b3)/(a2+ab+b2)
  =(a2+b2+ab)(a+b)/(a2+b2+ab)
  = a+b
  =40+31
  =71
• 11 years agoHelpfull: Yes(1) No(6)
• let a =40
  b=31
  then
  (a^3-b^3)/(a^2+ab+b^2)
  (a-b)(a^2+b^2+ab)/(a^2+ab+b^2)= a-b
  40 - 31
  9
  
• 11 years agoHelpfull: Yes(1) No(0)
• @SINDHURA
  
  bt still u are wrong because formula of
  
  a^3 + b^3 = (a + b)/(a^2 + b^2 -ab) bt u hav written (a + b)/(a^2 + b^2 + ab)
  
  :)
• 11 years agoHelpfull: Yes(1) No(1)
• sorry...
  
  a^3 + b^3 = (a +b)*(a^2 + b^2 -ab)
• 11 years agoHelpfull: Yes(1) No(1)
• ans:b)9
  let a=40 & b=31
  now, (a^3-b^3)/(a^2+b^2+ab)
  ={(a-b)*(a^2+b^2+ab)}/(a^2+b^2+ab)
  =(a-b)
  =9
• 11 years agoHelpfull: Yes(0) No(1)
• oh there is minus over there .......
  
• 11 years agoHelpfull: Yes(0) No(0)
• 9...
  apply a^3-b^3 formula
• 11 years agoHelpfull: Yes(0) No(0)
• (40^3)-(31^3)=(40-31)(40^2+40*31+31^2)
  so ans is 9
• 11 years agoHelpfull: Yes(0) No(0)
• we will use a^3 - b^3
  therefore:
  the numerical can be written as 40^3 - 31^3 = (40-31)*(40^2 + 40*31 + 31^2)/(40^2 + 40*31 + 31^2)
  hence 40-31 = 9
• 11 years agoHelpfull: Yes(0) No(0)
• APPLY BODMAS
  ANS IS 9
  
• 11 years agoHelpfull: Yes(0) No(0)