Using Cofactors of elements of third column, evaluate `Delta=|[1 , x, yz],[1, y, zx],[1, z, xy]|`

Explanation

`Delta=|{:(1,x,yz),(1,y,zx),(1,z,xy):}|`
Cofactors of the elements of third row
`A_(13)=(-1)^(1+3)|{:(1,y),(1,z):}|=z-y`
`A_(23)=(-1)^(2+3)|{:(1,x),(1,z):}|=(z-x)=x-z`
`A_(33)=(-1)^(3+3)|{:(1,x),(1,y):}|=y+x`
`therefore" "Delta=a_(13)A_(13)+a_(23)A_(23)+a_(33)A_(33)`
`=yz(z-y)+zx(x-z)+xy(y-x)`
`=yz^(2)-y^(2)z+zx^(2)-z^(2)x+xy^(2)-x^(2)y`