Generalized problem a + b + c = m, a^2 + b^2 + c^2 = m^2/3 2 ( a b + b c + c a) = ( a + b + c) ^2 - ( a^2 + b^2 + c^2) = m^2 - m^2/3 2(a^2 + b^2 + c^2) - (a b + b c +c a) = 2 m^2/3 - 2 m^2 /3 = 0 ( a - b) ^2 + ( b - c) ^2 + ( c - a) ^2 = 0 a = b = c = ( a + b + c) /3 = m /3 Hereby a^(2021) + b^ (2022) + c^ (2023) = ( m /3) ^ (2021) * ( 1 + a + a ^2)
Good one.
1
It's directly visible, all variables are 1. But I like your approach, it teaches other concepts as well.
Generalized problem
a + b + c = m,
a^2 + b^2 + c^2 = m^2/3
2 ( a b + b c + c a)
= ( a + b + c) ^2 - ( a^2 + b^2 + c^2)
= m^2 - m^2/3
2(a^2 + b^2 + c^2) - (a b + b c +c a)
= 2 m^2/3 - 2 m^2 /3
= 0
( a - b) ^2 + ( b - c) ^2 + ( c - a) ^2
= 0
a = b = c = ( a + b + c) /3 = m /3
Hereby
a^(2021) + b^ (2022) + c^ (2023)
= ( m /3) ^ (2021) * ( 1 + a + a ^2)
3
One could guess just by looking that the solution of both equations is 1
Ans=1
1+1+1=3
1^2+1^2+1^2=3
1^2021+1^2022+1^203=3
My goodness!! Is it the way to solve this problem? You will get hardly 5 seconds for this
A=b=c= 1
So last answer is 3