SOLUTION: a+b+c=3 and a^2 + b^2 +c^2 =3 solve for a^2022 +b^2022 +c^2022 show proof

I will start the solution from proving another statement


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    |   If 3*(a^2 + b^2 + c^2) = (a+b+c)^2,        |
    |   where "a", "b" and "c" are real numbers,   |
    |            then a = b = c.                   |
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Proof


From 3*(a^2+b^2+c^2) = (a+b+c)^2,  you can easily deduce making FOIL, that

    2a^2 + 2b^2 + 2c^2 = 2ab + 2ac + 2bc.    (1)


Next, you can re-write and transform (1) in this way

    (a^2-2ab+b^2) + (a^2-2ac+c^2) + (b^2-2bc+c^2) = 0,

or, equivalently,

    (a-b)^2 + (a-c)^2 + (b-c)^2 = 0.          (2)


But the squares of real numbers are always non-negative, and the sum of non-negative numbers 
can be equal to 0 (zero) if and only if each of these non-negative numbers is zero.

So, we conclude from (2) that  a = b = c,  and the statement is PROVED.


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    |   Now, let's apply this statement to our case.   |
    +--------------------------------------------------+


We are given that a+b+c = 3  and  a^2+b^2+c^2 = 3.

So, in our case  3*(a^2 + b^2 + c^2) = 3*3 = 9 = 3^2 = (a + b + c)^2.


Thus we can apply this theorem, proved above, to conclude that a = b = c.


Next, since a+b+c = 3, it means that  a = b = c = 1.


And finally, we obtain  a^2022 + b^2022 + c^2022 = 1^2022 + 1^2022 + 1^2022 = 1 + 1 + 1 = 3.


ANSWER.  Under given conditions,  a^2022 + b^2022 + c^2022 = 3.