. (40*40*40 – 31*31*31)/(40*40+40*31+31*31)=?
a)8
b)9
c)71
d)51
Solution: a3 -b3 =(a-b)*(a2+a*b+b2) so from this formula we will find (a-b) value

• Answer is 9.
  let a=40,b=31
  so, (40^3-31^3)/(40^2+40*31+31^2)
  Now as we know, a^3-b^3=(a-b)*(a^2+a*b+b^2)
  (40-31)(40^2+40*31+31^2)/(40^2+40*31+31^2)
  So, (40-31)=9.
• 10 years agoHelpfull: Yes(12) No(0)
• Let, a=40 and b=31
  (a^3-b^3)/(a^2+ab+b^2)=(a-b)=(40-31)=9
  Option b)9
• 10 years agoHelpfull: Yes(1) No(1)
• let a=40,b=31,
  (a^3-b^3)/(a^2+ab+b^2)=(a-b)(a^2+ab+b^2)/(a^2+ab+b^2)=(a-b)=40-31=9
  b)9
• 10 years agoHelpfull: Yes(0) No(0)
• the solution for this kind of questions is
  (a+b) i.e. a-40, b=31
  therefore, (40+31)=71
  options c is correct
• 10 years agoHelpfull: Yes(0) No(1)
• a^3 - b^3 = (a - b)(a^2 + a*b +b^2)
  40^3 - 31^3 = (40 - 31)(40^2 + 40*31 +31^2)/(40^2 + 40*31 +31^2) = 9
• 10 years agoHelpfull: Yes(0) No(0)
• let a=40 and b=31
  simply the above formula saatisfies the alzebric formula
  so a-b=40-31=9
• 10 years agoHelpfull: Yes(0) No(0)
• Trick- Guys try to solve these type of question by solving the last digit.
  takink the last digit of 40 i.e 0
  , likewise last digit of 31 is 1,
  Now (0*0*0-1*1*1)/(0*0+0*1+1*1)
  =(0-1)/(1)
  Now for numerator clearly (0-1) =9 i.e the last digit of the Numerator.
  Denominator is 1
  So, the ans will will be defenitly will the last digit 9.
  Only one option satsfy so ans is B.
• 10 years agoHelpfull: Yes(0) No(0)