Cubic system
⎩ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎧ 3 a = ( b + c + d ) 3 3 b = ( c + d + e ) 3 3 c = ( d + e + a ) 3 3 d = ( e + a + b ) 3 3 e = ( a + b + c ) 3
a , b , c , d , e are real numbers satisfying the above system. Determine the sum of all possible values of a .
The answer is 0.
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2 solutions
Note that if ( a , b , c , d , e ) is a solution, then so is ( − a , − b , − c , − d , − e ) . Thus, the sum of all possible values of a must be 0 .
W.L.O.G suppose a ≥ b ≥ c ≥ d ≥ e , note that e = 3 ( a + b + c ) 3 ≥ 3 ( d + b + c ) 3 = a e ≥ a This conclusion forces a = b = c = d = e . Therefore, we just need to solve 3 a = ( 3 a ) 3 , which gives a = − 3 1 , 0 , 3 1 .