Using properties of determinants prove the following: |3x -x+y -x+z| |x-y 3y -y-z| |-y-z 3z| = 3(x+y+z)(xy+yz+xz)

First take LHS: let Δ=|3x -x+y -x+z| |x-y 3y -y-z| |-y-z 3z|[Applying C1→C1+C2+C3]=|x+y+z -x+y -x+z| |x+y+z 3y -y-z| |x+y+z 3z|Taking x + y + z common from C1=(x+y+z)|1 -x+y -x+z| |1 3y -y-z| |1 3z|Applying R1→R1-R3, R2→R2-R3=(x+y+z)|0 -x+y -x+z| |0 3y -y-z| |1 3z|Expanding along C1=(x+y+z)(1×|-x+y -x+z| | 3y -y-z|) =(x+y+z)[(-x+z)(-y-z)-(3y)(-x+z)]=(x+y+z)(3xy+3yz+3zx)=3(x+y+z)(xy+yz+zx)LHS = RHS Hence proved.