Question

Verify the following

$\left(i\right)(ab+bc)(ab-bc)+(bc+ca)(bc-ca)+(ca+ab)(ca-ab)=0$

Answer 1

$L.H.S.:$

$(ab+bc)(ab-bc)+(bc+ca)(bc-ca)+(ca+ab)(ca-ab)$

$=(ab{)}^2-(bc{)}^2+(bc{)}^2-(ca{)}^2+(ca{)}^2-(ab{)}^2$

[Using identity, $(a+b)(a-b)=(a^2-b^2)$]

$=(a^2{b}^2-b^2{c}^2)+(b^2{c}^2-c^2{a}^2)+(c^2{a}^2-a^2{b}^2)$

$=a^2{b}^2-b^2{c}^2+b^2{c}^2-c^2{a}^2+c^2{a}^2-a^2{b}^2$

$=0$

$=R.H.S.$
Hence, verified.