Proving an algebraic identity

Question

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:

Answer 1

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.

Answer 2

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.

Answer 3

$$(a+b)(b+c)(c+a)$$ $$=(ab+bc+ca+b^2)(c+a)$$ $$=(ab+bc+ca)(a+c)+ab^2+b^2c$$ Adding $abc$ gives: $$(ab+bc+ca)(a+c)+ab^2+b^2c+abc$$ $$=(ab+bc+ca)(a+c)+b(ab+bc+ac)$$ $$=(ab+bc+ca)(a+b+c)$$

Answer 4

$(a+b+c)(ab+bc+ca)−abc$

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

Answer 5

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.