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[ax-by-cz,ay+bx,az+cx],[bx+ay,by-cz-ax,b...

[ax-by-cz,ay+bx,az+cx],[bx+ay,by-cz-ax,bz+cy],[cx+az,ay+bz,cz-ax-by],[=,(a^(2)+b^(2)+c^(2))(ax+by+cz)(x^(2)+y^(2)+z^(2))]

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Prove the following: |[ax-by-cz,ay+bx,az+cx],[bx+ay,by-cz-ax,bz+cy],[cx+az,ay+bz,cz-ax-by]| = (a^2+b^2+c^2)(ax+by+cz)(x^2+y^2+z^2)

det[[ Prove that ax-by-cz,ay+bx,cx+azay+bx,by-cz-ax,bz+cycx+az,bz+cy,cz-ax-by]]=(x^(2)+y^(2)+z^(2))(a^(2)+b^(2)+c^(2))(ax+by+cz)

If ax+cy+bz=X, cx+by+az=Y, bx+ay+cz=Z, show that (a^(2)+b^(2)+c^(2)-bc-ca-ab)(x^(2)+y^(2)+z^(2)-yz-zx-xy)=X^(2)+Y^(2)+Z^(2)-YZ-ZX-XY

Given x=cy+bz,y=az+cx and z=bx+ay, then prove a^(2) +b^(2) +c^(2) +2abc =1.

Given x=cy+bz,y=az+cx and z=bx+ay, then prove a^(2) +b^(2) +c^(2) +2abc =1.

Given x=cy+bz,y=az+cx and z=bx+ay, then prove a^(2) +b^(2) +c^(2) +2abc =1.

Given x=cy+bz,y=az+cx and z=bx+ay, then prove a^(2) +b^(2) +c^(2) +2abc =1.

If x=cy+bz,y=cx+az,z=bx+ay the value of a^(2)+b^(2)+c^(2)-1 is (A)abc(B) abc (C)2abc(D)-2abc

If x@+y^(2)+z^(2)!=0,x=cy+bz,y=az+cx, and z=bx+ay, then a^(2)+b^(2)+c^(2)+2abc=