If `(1)/(root(3)(4) + root(3)(2) + 1) = xroot(3)(4) + y root(3)(2) + z`, where x, y and z are rational, then find the value of x,y and z

Explanation

Let `root(3)(4) = a, root(3)(2) =b and 1 = c`
Then the given expression `= (1)/(a + b + c)`
`= (1)/(a + b + c) xx (a^(2) + b^(2) + c^(2) - ab - ab - ca)/(a^(2) + b^(2) + c^(2) - ab - bc - ca)`
`= (a^(2) + b^(2) + c^(2) - ab - bc - ca)/(a^(3) + b^(3) + c^(3) - 3abc)`
`= ((root(3)(4))^(2) + (root(3)(2))^(2) + (1)^(2) - root(3)(4). root(3)(2) - root(3)(2) xx 1 - 1 xx root(3)(4))/((root(3)(4))^(3) + (root(3)(2))^(3) + (1)^(3) - 3 root(3)(4).root(3)(2).1)`
`= (root(3)(16) + root(3)(4) + 1- root(3)(8) - root(3)(2) - root(3)(4))/(4 + 2 + 1 - 3 root(3)(8))`
`= (2root(3)(2) + 1 -2-root(3)(2))/(7 - 3 xx 2) = (root(3)(2) -1)/(7 -6)`
`= (root(3)(2) -1)/(1) = root(3)(2) -1`
As per question, `root(3)(2) - 1 = x root(3)(4) + yroot(3)(2) + z`
`:. x = 0, y =1 and z = -11`