Question

Find the value of the expression in terms of $a,b,$ and $c$.
$(ab+bc)(-cb+ba)+(bc+ca)(-ac+cb)+(ca+ab)(ac–ba)$

• A. $a+b+c$
• B. $3ab$
• C. $3b$
• D. None of above

Answer 1

The correct option is D None of above

$(ab+bc)(-cb+ba)+(bc+ca)(-ac+cb)+(ca+ab)(ac–ba)$

After arranging the expressions, we get,
$(ab+bc)(ab–bc)+(bc+ca)(bc–ca)+(ca+ab)(ca–ab)$

Now, using the identity, $(a+b)(a-b)=a^2-b^2$, the above expression become,

= [${\left(ab\right)}^2$-${\left(bc\right)}^2$]+[ ${\left(bc\right)}^2$-${\left(ca\right)}^2$]+[ ${\left(ca\right)}^2$-${\left(ab\right)}^2$]

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

= $0$