# Solving unusual equation`[[x,1,2],[2,x,1],[1,2,x]]=0`

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You need to evaluate the determinant of the given matrix.

`[[x,1,2],[2,x,1],[1,2,x]] ` = `x*x*x + 2*2*2 + 1*1*1 - 2*x*1 - 2*x*1 - 2*x*1`

Adding the terms yields:

`[[x,1,2],[2,x,1],[1,2,x]] = x^3 +9 - 6x`

`[[x,1,2],[2,x,1],[1,2,x]] = 0 =gt x^3 - 6x + 9 = 0`

The roots of the equation above may be searched between the divisors of 9.

`D_9 = {+-1 ; +-3 ; +-9}`

Put `x = 3 =gt 27 - 18 + 9 != 0`

Put `x =-3 =gt -27 + 18 + 9 = 0 =gt x= -3` is a root of `x^3 - 6x + 9 = 0.`

Write the reminder theorem:

`x^3 - 6x + 9 = (x+3)(ax^2 + bx + c)`

To find the next two roots, you need to find the quadratic`ax^2 + bx + c.`

Removing the brackets yields:

`x^3 - 6x + 9 = ax^3 + bx^2 + cx + 3ax^2 + 3bx + 3c`

Equating the coefficients of the terms having the same power yields:

a = 1

3a+b = 0 => b = - 3 3c = 9 => c = 3

The quadratic is `x^2 - 3x + 3 = 0`

Solving the quadratic yields: `x_(2,3) = (3+-sqrt(9-12))/2`

`x_(1,2) = (3+-sqrt(-3))/2 =gt x_(2,3) = (3+-i*sqrt(3))/2`

**The solutions to the given equation are one real and two complex: `x = -3` and `x_(2,3) = (3+-i*sqrt(3))/2` **