You need to evaluate the determinant of matrix of coefficients of variables x,y,z, such that:

`Delta = [(k,1,1),(1,k,1),(1,1,k)]`

`Delta = k^3 + 2 - 3k`

If `Delta!=0` , hence, the system has an uniques solution, such that:

`Delta!=0 => k^3 + 2 - 3k != 0 => k^3 + 3 - 1 - 3k != 0`

You need to group the terms, such that:

`(k^3 - 1) + (3 - 3k) != 0`

Converting the difference of cubes into a product, yields:

`(k - 1)(k^2 + k + 1) - 3(k - 1) != 0`

Factoring out `(k-1)` yields:

`(k - 1)(k^2 + k - 2) != 0 => {(k-1!=0),(k^2 + k - 2!=0):}`

`{(k!=1),(k_(1,2)!=(-1+-sqrt(1+8))/2)):}`

`(k!=1),(k_(1,2)!=(-1+-3)/2) =>(k!=1),(k_1!=1, k_2!=-2)`

**Hence, evaluating the values of k for the system to have an unique solution yields **`k!=-2, k!=1.`