If `x+y+z=0` , prove that `|x a y b z c y c z a x b z b x c y a|=x y z|a b cc a bb cc|`

Answer

N/a

Explanation

Since `x+y+z=0` also we have to prove
`|(xa,yb,zc),(yc,za,xb),(zb,xc,ya)|=xyz|(a,b,c),(c,a,b),(b,c,a)|`
`:. LHS =|(xa,yb,zc),(yc,za,xb),(zb,xc,ya)|`
`=xa(za.ya-xb.xc)-ub(yc.ya-xb.b)+zc(yc.xc-za-zb)`
`=xa(a^(2)yz-x^(2)bc)-yb(y^(2)ac-b^(2)xz)+zc(c^(2)xy-z^(2)ab)`
`=xyza^(3)-x^(3)abc-y^(3)abc+b^(3)xyz+c^(3)xyz-z^(3)abc` `=x yz(a^(3)+b^(3)+c^(3))-abc(x^(3)+y^(3)+z^(3))`
`=x yz (a^(3)+b^(3)+c^(3))-abc(3x yz)`
`[ :' x+y+z=0impliesx^(3)+y^(3)+z^(3)-3xyz]`
`=x yz(a^(3)+b^(3)+c^(3)-3abc)`............i
Now `RHS=x yz|(a,b,c),(c,a,b),(b,c,a)|=xyz|(a+b+c,b,c),(a+b+c,a,b),(a+b+c,c,a)|[ :'C_(1)toC_(1)+C_(2)+C_(3)]`
`=x yz(a+b+c)|(1,b,c),(1,a,b),(1,c,a)|` [taking `(a+b+c)` common from `C_(1)`]
`=x yz(a+b+c)|(0,b-c,c-a),(0,a-c,b-a),(1,c,a)|`
Expanding along `C_(1)` `[ :' R_(1)toR_(1)-R_(3)` and `R_(2)toR_(2)-R_(3)]`
`=x yz(a+b+c)[(b-c)(b-a)-(a-c)(c-a)]`
`=x yz(a+b+c)(b^(2)-ab-bc+ac+a^(2)+c^(2)-2ac)`
`=x yz (a+b+c)(a^(2)+b^(2)+c^(2)-ab-bc-ca)`
`=xyz(a^(3)+b^(3)+c^(3)-3abc)`.................ii
From Eqs i and ii
LHS `=` RHS
`implies |(xa,yb,zc),(yc,za,xb),(zb,xc,ya)|=x yz|(a,b,c),(c,a,b),(b,c,a)|` Hence proved.