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[1552],[|[x,a],[y,b],[z,c]|+b|=0]...

[1552],[|[x,a],[y,b],[z,c]|+b|=0]

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Without expanding the determinant prove the following. |[x,a,x+a],[y,b,y+b],[z,c,z+c]|=0

|[x,a,x+a],[y,b,y+b],[z,c,z+c]|=0

Show without expanding at any stage that: [x,a,x+a],[y,b,y+b],[z,c,z+c]|=0

Using the property of determinants and without expanding {:[( x,a,x+a),( y,b,y+b),(z,c,z+c)]:} =0

Using the property of determinants and without expanding {:|( x,a,x+a),( y,b,y+b),(z,c,z+c)|:} =0

Using the property of determinants and without expanding {:|( x,a,x+a),( y,b,y+b),(z,c,z+c)|:} =0

Using the property of determinants and without expanding {:[( x,a,x+a),( y,b,y+b),(z,c,z+c)]:} =0

If x+y+z=0 prove that |[a x, b y, c z],[ c y, a z ,b x],[ b z ,c x ,a y]|=x y z|[a, b, c],[c ,a ,b],[b ,c ,a]|

If x+y+z=0 prove that |[a x, b y, c z],[ c y, a z ,b x],[ b z ,c x ,a y]|=x y z|[a, b, c],[c ,a ,b],[b ,c ,a]|

|(x,a,x+a),(y,b,y+b),(z,c,z+c)| = 0