Solve the equation `|[a-x, c, b], [ c, b-x, a], [b, a, c-x]|=0` where `a+b+c!=0.`

Answer

`x=a+b+c,ne sqrt((1)/(2)(a-b)^(2)+(b-c)(2)+(c-a)^(2))`

Explanation

Applying `C_(1) to C_(1) +C_(2)+C_(3)` we get
`Delta=|{:(a+b+c-x,,c,,b),(a+b+c-x,,b-x,,a),(a+b+c-x,,a,,c-x):}|`
`=(a+b+c-x) |{:(1,,c,,b),(1,,b-x,,a),(1,,a,,c-x):}|`
Applying `R_(1) to R_(1)-R_(2),R_(2)toR_(2)-R_(3),` we get
`Delta =(a+b+c-x) |{:(0,,c-b+x,,b-a),(0,,b-a-x,,a-c+x),(1,,a,,c-x):}|`
`=(a+b+c -x){(x^(2)+x(a-b)+c-b)(a-c)`
`+(x+a-b)(b-a)`
`=(a+b+c-x){(x^(2)+x(a-b)+(c-b)(a-c)`
`+x(b-a)-(a-b)^(2)}`
`=(a+b+c-x)(x^(2)+ac-c^(2)-ab+bc-a^(2)-b^(2)+2ab)`
`=(a+b+c-x)(x^(2)-a^(2)-b^(2)-c^(2)+ab+bc+ca)`
`=(a+b+c-x)(x^(2)-(1)/(2){(a-b)^(2)+(b-c)^(2)+(c-a)^(2))})`
Since `Delta =0` we have
`x=a +b +c, +- sqrt((1)/(2)((a-b)^(2)+(b-c)^(2)+(c-a)^(2)))`