Question

Simplify this. Answer should be (a+b)(b+c)(c+a)

Answer 1

We know that $p^3+q^3+r^3=3pqr$ when $p+q+r=0$.
Let's simplify the numerator first,
[tex]\begin{array}{ccc} (a^2-b^2)^3+(b^2-c^2)^3+(c^2-a^2)^3&=&p^3+q^3+r^3\\ &=&3pqr\\ &=&3(a^2-b^2)(b^2-c^2)(c^2-a^2) \end{array}[/tex]
And the denominator is,
[tex]\begin{array}{ccc} (a-b)^3+(b-c)^3+(c-a)^3&=&p^3+q^3+r^3\\ &=&3pqr\\ &=&3(a-b)(b-c)(c-a) \end{array}[/tex]
Now the complete fraction gives us,
$(a+b)(b+c)(c+a)$