evaluate: `|(x+4,x,x),(x,x+4,x),(x,x,x+4)|`

Answer

N/a

Explanation

We have `|(x+4,x,x),(x,x+4,x),(x,x,x+4)|=|(2x+4,2x+4,2x),(x,x+4,x),(x,x,x+4)|` `[ :'R_(1)toR_(1)+R_(2)]`
`=|(2x,2x,2x),(x,x+4,x),(x,x,x+4)|+|(4,4,0),(x,x+4,x),(x,x,x+4)|`
[ here given determinant is expressed in sum of two determinants]
`=2x|(1,1,1),(x,x+4,x),(x,x,x+4)|+4|(1,1,0),(x,x+4,x),(x,x,x+4)|`
[Taking `2x` common from list row of first determinant and 4 from now of second determinant]
Applying `C_(1)toC_(1)-C_(3)` and `C_(2)toC_(2)-C_(3)` in first and applying `C_(1)toC_(1)-C_(2)` is second, we get
`=2x|(0,0,1),(0,4,x),(-4,-4,x+4)|+4|(0,1,0),(-4,x+4,x),(0,x,x+4)|`
Expanding both the along first column we get
`2x[-4(-4)]+[4(x+4-0)]`
`=2x xx16+16(x+4)`
`=32x+16x+64`
`=16(3x+4)`