Question

S=93^3-57^3-36^3 is divisiable by which number

Answer 1

we know
when ,
a+b+c=0
then,
a^3+b^3+c^3=3abc
so,
93^3+(-57)^3+(-36)^3=3(93)(-57)(-36)
hence 93^3-57^3-36^3 divisible by 3(93)(-57)(-36)