If `x + y + z = 0,` prove that `|(ax,by,cz),(cy,az,bx),(bz,cx,ay)|=xyz|(a,b,c),(c,a,b),(b,c,a)|`

Explanation

since the determinant is symmmetrical it is easier to expand by Sarrus rule.
` |{:(ax,,by,,cz),(cy ,,az,,bx),(bz,,cx,,ay):}|`
`= xyz (a^(3)+b^(3)+c^(3)) - abc (x^(3) +y^(3)+z^(3))`
`= xyz (a^(3) +b^(3) +c^(3) -3 abc)`
`-abc (x^(3) +y^(3) +z^(3) -3xyz)`
`=xyz ( a^(3) +b^(3) +c^(3) -3abc) -abc (x+y+z)`
` xx (x^(2) +y^(2) +z^(2) -xy -yz -zx)`
` =xyz (a^(3) +b^(3) +c^(3)-3abc)`
`(:' x+y + z=0)`
` =xyz|{:(a,,b,,c),(c ,,a,,b),(b,,c,,a):}|`