Proving an algebraic identity Prove:
$$(a + b + c)(ab + bc + ca) - abc = (a + b)(b + c)(c + a)$$
Problem:
I am not sure how to proceed after expanding the brackets on the RHS. I am not sure if I also expanded correctly. My solution is:
 A: Try putting $a+b+c=x$. Then consider
\begin{align}
(x-a)(x-b)(x-c) &= (x-a)(x^{2}-(b+c)x+bc)\\
                &= x^{3}-(b+c)x^{2}+bcx-ax^{2}+ax(b+c)-abc\\
                &= x^{3}-x^{2}(a+b+c)+x(ab+bc+ac)-abc\\
\end{align}
note that
$$x^{3}-x^{2}(a+b+c)=0$$
A little rearrangement will give you the identity you seek.
A: Both sides are homogeneous polynomials in $a,b,c$ with degree $3$. You may check they agree at $1,1,1$ and that the LHS vanishes at $a+b=0$, $a+c=0$, $b+c=0$ (to check one condition is enough, by symmetry) to have they are the same polynomial. But if $a+b=0$
$$(a+b+c)(ab+ac+bc)-abc = c(ab)-abc =0$$
and we are done.
A: $$(a+b)(b+c)(c+a)$$
$$=(ab+bc+ca+b^2)(c+a)$$
$$=(ab+bc+ca)(a+c)+ab^2+b^2c$$
Adding $abc$ gives:
$$(ab+bc+ca)(a+c)+ab^2+b^2c+abc$$
$$=(ab+bc+ca)(a+c)+b(ab+bc+ac)$$
$$=(ab+bc+ca)(a+b+c)$$
A: $(a+b+c)(ab+bc+ca)−abc$
P1
$f(a) = (a+b+c)(ab+bc+ca)−abc$
$f(-b) = c(-b^2 + bc -bc) - cb^2 = 0$
Like wise symmetrically
a+b,b+c,c+a are factors
You can use the symmetry argument and claim that these are the only factors
Or you can go down the path of expand and factoring again
A: Similar to the solutions of Jack D'Aurizio and sidt36:
Consider each side as a function of $a$:
$$L(x) = (x + b + c)(xb + bc + cx) - xbc$$
$$R(x) = (x + b)(b + c)(c + x)$$
They are both quadratic, and so can be expressed as $A(x-r)(x-s)$ (where $r,s$ in general may be complex).
But it is easily verified that $$L(-b) = R(-b) = 0\\L(-c) = R(-c) = 0\\L(0) = R(0) = bc(b+c)$$
from which we see that $r = -b, s = -c, A = b+c$ for both, and therefore they are equal.
