x = 21/3 + 22/3 ---> x3 - 6x - 6 = 0
First, let's evaluate: [ 21/3 + 22/3 ]3
Notice that [ a + b ]3 = a3 + 3a2b + 3ab2 + b3
The x3 part: [ 21/3 + 22/3 ]3 = [21/3]3 + 3·[21/3]2·[22/3] + 3·[21/3]·[22/3]2 + [22/3]3
= 23/3 + 3·[22/3]·[22/3] + 3·[21/3]·[24/3] + 26/3
= 2 + 3·24/3 + 3·25/3 + 4
= 6 + 3·2·21/3 + 3·2·22/3
= 6 + 6·21/3 + 6·22/3
The 6x part: 6·[ 21/3 + 22/3 ] = 6·21/3 + 6·22/3
Thus: x3 - 6x - 6 = [ 6 + 6·21/3 + 6·22/3 ] - [ 6·21/3 + 6·22/3 ] - 6
= 6 + 6·21/3 + 6·22/3 - 6·21/3 - 6·22/3 - 6
= 0