Amar! Akbar! Anthony!

Let the ages of Amar, Akbar and Anthony be $x, y$ and $z$ respectively.

Hence the sum of their present ages is $x+y+z=80$

Three years ago the ages of Amar, Akbar, Anthony are $(x-3),(y-3),(z-3)$ respectively.

Thus three years ago. the sum of their age is $(x+y+z)-9$ , but $x+y+z=80$ .

Therefore, the sum of their ages three years ago is $80-9=71$ .

Required sum = (80 - 3 x 3) years = (80 - 9) years = 71 years.

I ignored the assumption and took 80-(3x3)=71

Let the ages of Amer,Akbar & Anthony $x,y,z$ respectectly

The sum of their ages is $x+y+z=80$

So the sum of their ages three years ago is ((x-3)+(y-3)+(z-3)=x+y+z-(3+3+3)=80-9=\boxed{\large{71}}\

80 - 3x3=71

x + y + z = 80 (x-3) + (y-3) + (z-3) = x + y + z - 9 = 80 - 9 = 71

a+b+c=80 (a-3)+(b-3)+(c-3)=x x=71 smile emoticon

Three years ago, each of them will be 3 years younger, or will have 'lost' 3 years. 80 - (3 x 3) = 80 - 9 = 71