Trivial cubic sum

Consider $f(a) = a(b-c)^3 + b(c-a)^3 + c(a-b)^3$

We can check that $f(-(b+c))=0$

Hence, by the remainder-factor theorem, we know that $(a+b+c)$ is a factor of $f(a)$

Let's assume that $f(a) = (a+b+c)\cdot (A)$ for a bunch of terms $(A)$ (let's not care about what are the terms)

We are given that $a+b+c = 0$

Hence, the only possible value of $f(a)$ under the condition that $a+b+c = 0$ is...

$f(a) = 0 \cdot (A) = \Large{\fbox{0}}$

Setting $a=b$ makes the expression $a(b-c)^3+b(c-a)^3+c(a-b)^3$ zero. $a-b$ is therefore a factor. Since the expression is cyclic $b-c$ and $c-a$ are also factors.

However, $(a-b)(b-c)(c-a)$ is of degree 3 and $a(b-c)^3+b(c-a)^3+c(a-b)^3$ is of degree 4. A cyclic factor of degree 1 is needed. It follows that $a(b-c)^3+b(c-a)^3+c(a-b)^3=(a-b)(b-c)(c-a)(a+b+c)$ , but $a+b+c=0$ . The only value the expression takes is 0 and hence this is the maximum value.

Note that a-b is a root. By symmetry, so are b-c, and c-a. Factor the expression, which neatly becomes (a-b)(b-c)(c-a)(a+b+c). But the last factor is simply 0, so 0 is the only possible value.

Since setting $a=b$ , $b=c$ , or $c=a$ in the equation all cause it to evaluate to $0$ , $(a-b)$ , $(b-c)$ , and $(c-a)$ all divide the expression.

Note that both the original expression and the $(a-b)(b-c)(c-a)$ are symmetric; that is, if they are set as functions $f(a,b,c)$ , $f(a,b,c)=f(b,c,a)=f(c,a,b)$ . Also, the original expression is cubic in all variables, while $(a-b)(b-c)(c-a)$ is quadratic. Therefore the factor by which $(a-b)(b-c)(c-a)$ needs to be multiplied must be both linear and symmetric.

Since $(a+b+c)$ is the only possible linear symmetric polynomial, $a(b-c)^3+b(c-a)^3+c(a-b)^3=(a-b)(b-c)(c-a)(a+b+c) * k$ , where k is some real number. Substituting in $a+b+c=0$ shows that the original expression must identically equal $0$ everywhere, so the maximum is $\boxed{0}$ .

Since $a(b-c)^3+b(c-a)^3+c(a-b)^3=(a-b)(a-c)(c-b)(a+b+c)$ and $a+b+c=0$ , the expression is always equals to $0$ , the largest possible value of it is also $0$ .

First, we expand one of the terms in the expression. $a^{}(b-c)^{3} = a^{}b^{3} - 3a^{}b^{2}c^{} + 3a^{}b^{}c^{2} - a^{}c^{3}$

Using similar expansion for the other 2 terms and summing them, we have $a^{}(b-c)^{3} + b^{}(c-a)^{3} + c^{}(a-b)^{3}$

= $a^{}b^{3} - 3a^{}b^{2}c^{} + 3a^{}b^{}c^{2} - a^{}c^{3}$

$+ b^{}c^{3} - 3b^{}c^{2}a^{} + 3b^{}c^{}a^{2} - b^{}a^{3}$

$+ c^{}a^{3} - 3c^{}a^{2}b^{} + 3c^{}a^{}b^{2} - c^{}b^{3}$

Canceling terms, this equals $a^{}b^{3} + b^{}c^{3} + c^{}a^{3} - a^{}c^{3} - b^{}a^{3}- c^{}b^{3}.$

With factorisation, it can be rewritten as $a^{}b^{}(b^{2} - a^{2}) + b^{}c^{}(c^{2} - b^{2}) + c^{}a^{}(a^{2} - c^{2})$ = $ab(b+a)(b-a) + bc(c+b)(c-b) + ca(a+c)(a-c)$

Now we employ the fact that $a+b+c = 0$ . It gives us $a+b = -c, b+c = -a, c+a = -b$

Substituting this into the expression gives

$ab(-c)(b-a) + bc(-a)(c-b) + ca(-b)(a-c)$

= $( -abc )( b-a+c-b+a-c) = (-abc)(0) = 0$

Thus the expression has only one possible value , $0$ .

Using AM GM inequality we have:

$a(b-c)^3 + b(c-a)^3 +c(a-b)^3 \le \frac{a + (b-c)^3 + b + (c-a)^3) + c + (a-b)^3)}{4}$

that (with some algebra and remembering that $a + b + c = 0$ ) is:

$\frac{3}{4} a(b^2-c^2) + b(c^2 - a^2) + c(a^2 - b^2)$

Than, applying again AM-GM inequality we have:

$\le \frac{3}{16}( a + b^2 -c^2 +b +c^2 -a^2 +c+a^2-b^2) = 0$

Since the initial expression evaluated for $a = b = c = 0$ is 0, it is the maximum value.

a(b-c)^3+b(c-a)^3+c(a-b)^3
=a(b-c)^3+b(c-a)^3-(a+b)(a-b)^3
=a{(b-c)^3-(a-b)^3}+b{(c-a)^3-(a-b)^3}
=a{(3b)^3-3(b-c)(a-b)}+b{(3a)^3+3(c-a)(a-b)3a}
=27ab^3-27a^3b+9ab{(c-a)(a-b)-(a-b)(b-c)}
=27ab(b+a)(b-a)-27ab(b-a)
=54abc(a-b)....(1)

now from A.M-G.M ineq. (a+b+c)^3/27>=abc....as a+b+c=0 so 27abc<=0...so the highest value of 27abc is o.. hence we can interpret that the highest value of eqtn (1) is 0...