If x+y+z=0=a+b+c, then `|{:(xa,yd,zc),(yc,za,xb),(zb,xc,ya):}|=`

If x+y+z=0=a+b+c, then |{:(xa,yb,zc),(yc,za,xb),(zb,xc,ya):}|=

If x+y+z = 0 prove that |{:(xa,yb,zc),(yc,za,xb),(zb,xc,ya):}|=xyz|{:(a,b,c),(c,a,b),(b,c,a):}|

If x+y+z=0 and |(xa, yb, zc),(yc,za, xb),(zb,xc, ya)|=k|(a,b,c),(b,c,a),(c,a,b)| then k =

If x+y+z=0 , prove that |[xa, yb,zc],[yc, za, xb], [zb, xc, ya]|=x y z|[a, b, c],[c ,a, b],[b ,c,c]|

If x+y+z=0 , prove that |[xa, yb,zc],[yc, za, xb], [zb, xc, ya]|=x y z|[a, b, c],[c ,a, b],[b ,c,c]|

If x+y+z=0 prove that | xa yb zc yc za xb zb xc ya y|=x y z|a b c c a b b c a|

If a^xb=b^yc=c^za=1 then (xyz)^3=

Without expanding determinant show that |(xa,yb,zc),(a^2,b^2,c^2),(1,1,1)|=|(x,y,z),(a,b,c),(bc,ca,ab)|

If P=[{:(x,0,0),(0,y,0),(0,0,z):}] and Q=[{:(a,0,0),(0,b,0),(0,0 ,c):}] then prove that PQ=[{:(xa,0,0),(0,yb,0),(0,0,zc):}]=QP .