if a(b+c)=33
b(c+a)=40
c(a+b)=49
then find abc options are
14 square root 33
84
122
cannot be founded using given data

• a(b+c)=33
  so a=11 or 3
  b+c=11 or 3
  c(a+b)=49
  c=7
  a+b=7
  so a=3
  and b=4
  and c=7
  abc=84
• 8 years agoHelpfull: Yes(17) No(1)
• simple question just factorize given no
  33= 3*11 = a(b+c)
  40=8*5=b(c+a)
  49=7*7=c(a+b)
  from the last equation it is clear that c=7 which means b must be = 4 and a = 3
  therefore a*b*c = 7*4*3= 84.
• 8 years agoHelpfull: Yes(7) No(1)
• as given
  ab+ac=33 …eq1
  bc+ab=40…eq2
  ac+bc=49.…eq3
  
  than eq1-eq2
  bc-ac=7…eq4
  
  eq2-eq3
  ac-ab=9 …eq5
  
  eq3-eq1
  bc-ab=16…eq6
  
  now again
  eq2+eq6
  bc=28
  eq1+eq5
  ab=21
  eq3+eq4
  bc=28
  
  but yet we did not get ab, so put the value of bc in eq2 and ab=12
  now
  ab*bc*ac=28*21*12
  (abc)^2=7056
  abc=84
• 8 years agoHelpfull: Yes(5) No(0)
• Put a=3,b=4,c=7
  3(4+7)=33
  4(7+3)=40
  7(3+4)=409
  
  So abc=3*4*7=84
• 8 years agoHelpfull: Yes(2) No(0)
• ANs......84
  a(b+c)= 33 => ab+ac=33-------(i)
  b(c+a)= 40 => bc+ab=40-------(ii)
  c(a+b)= 49 => ac+bc=49-------(iii)
  
  add above 3 eqn
  2ab+2bc+2ac= 122
  ab+ac+bc=61------(iv)
  
  Now solve eqn(i) and eqn (iv) we will get bc=28
  similarly ac=21 & ab=12
  
  Now ab*ac*bc= 5761
  abc= under root of 5761 that is 84
  
  ANswer is 84
• 8 years agoHelpfull: Yes(1) No(0)
• c(a+b)=49
  factors of 49=7*7
  c=7 and a+b=7
  then
  a(b+c)=33
  factors of 33=3*11
  so a=3 b=4 bcoz c=7
  answer =3*4*7=84
• 8 years agoHelpfull: Yes(0) No(0)
• a(b+c)=33=3*11
  b(c+a)=4*10
  c(a+b)=7*7
  a=3,b=4,c=7
  so,a*b*c=3*4*7=84
• 8 years agoHelpfull: Yes(0) No(0)