Simplify `(ab+bc-ca)^2`

Simplify: (a+b+c)(a^(2)+b^(2)+c^(2)-ab-bc-ca)

Simplify (2a^(2)b-3bc)-(a^(2)b+5bc-ca)

Verify that (ab+ bc) (ab- bc) + (bc+ ca) (bc- ca) + (ca + ab) (ca- ab) = 0

Simplify: (1)/(1+log_(a)bc)+(1)/(1+log_(b)ca)+(1)/(1+log_(c)ab)

Prove |[-bc, b^2+bc, c^2+bc] , [a^2+ac, -ac, c^2+ac] , [a^2+ab, b^2+ab, -ab]|=(ab+bc+ca)^2

Subtract 2(ab+bc+ca) " from " -ab-bc-ca .

Add the following expressions: ab + bc + ca and - bc - ca- ab

If ab + bc + ca = 0 , then the value of 1/(a^2 - bc) + 1/(b^2 - ca) + 1/(c^2 - ab) will be :

If ab+bc+ca=0 , then the value of ((b^(2)-ca)(c^(2)-ab)+(a^(2)-bc)(c^(2)-ab)+(a^(2)-bc)(b^(2)-ca))/((a^(2)-bc)(b^(2)-ca)(c^(2)-ab)) is