Question 1208296
.
a + b + c = 6
a² + b² + c² = 14, 
find the value of a⁸ + b⁸ + c⁸
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<pre>
It is assumed that a, b, c are integer numbers.


From  {{{a^2}}} + {{{b^2}}} + {{{c^2}}} = 14  we guess one solution

    a= 3,  b= 2,  c= 1.


Indeed,  then

    3 + 2 + 1 = 6

and

    3^2 + 2^2 + 1^2 = 9 + 4 + 1 = 14.


So, (a,b,c) = (3,2,1) is the basic solution.


There are 5 other solutions that are permutations of the basic triple.


But we should not worry about permutations, since they do not make influence on
the sum  {{{a^8}}} + {{{b^8}}} + {{{c^8}}}.


Also, it is clear from the equation for squares, that there are no other solutions in integer numbers.


Now we make direct calculation  {{{a^8}}} + {{{b^8}}} + {{{c^8}}} = {{{3^8}}} + {{{2^8}}} + {{{1^8}}} = 6818.


<U>ANSWER</U>.  Under given conditions, the sum  {{{a^8}}} + {{{b^8}}} + {{{c^8}}}  is  6818.
</pre>

Solved.


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