Let $f(a,b,c) = a(b-c)^3 + b(c-a)^3 + c(a-b)^3.$ Each solution of $f(a,b,c)=0$ has the form $f(a, a+p, a+q) =0,$ where $a\geq -10$ and $p, q$ are integers satisfying $0<p<q$ . From $f(a, a+p, a+q) = p q (q-p)(3a+p+q)$ and $0<p<q$ it follows that $- 3a = p + q,$ and hence $-10 \leq a \leq -1$ . For these values we have to add the number of solutions $p$ and $q$ , which is 75.

If we put $a=b$ then the value of expression is 0, $\Rightarrow (a-b)$ is a factor. Similarly, $(b-c)$ and $(c-a)$ should be the factor of the above expression. Since in the above expression all the terms are of degree 4 hence there must be a fourth factor of degree 1. $a(b-c)^3 + b(c-a)^3 + c(a-b)^3 = k(a-b)(b-c)(c-a)(a+b+c)$ . [Why do we know that the 4th factor must be $(a+b+c)$ ? This is because it is the only cyclic degree 1 polynomial. -Calvin] Now we need to find the value of $k$ by replacing $a=0,b=1,c=2$ , which gives $0(1-2)^3 + 1*(2-0)^3 + 2*(0-1)^3 = k(0-1)(1-2)(2-0)(0+1+2)$ Hence $0+8-2 = k(2*3) \Rightarrow k=1$ , so the above expression can be written as $(a-b)(b-c)(c-a)(a+b+c) = 0$ .

Since $(a-b)$ or $(b-c)$ or $(c-a)$ cannot be 0, hence $(a+b+c) = 0$ . Hence, we must have $a < 0$ , and we proceed to consider cases.

For $a = -10$ we get $b+c = 10$ . $b$ can take values from -9 onwards up to 4, hence 14 solutions.
For $a = -9$ we get $b+c = 9$ . $b$ can take values from -8 up to 4, hence 13 solutions.
For $a = -8$ we get $b+c = 8$ . $b$ can take values from -7 up to 3, hence 11 solutions.
For $a = -7$ we get $b+c = 7$ . $b$ can take values from -6 up to 3, hence 10 solutions.
For $a = -6$ we get $b+c = 6$ . $b$ can take values from -5 up to 2, hence 8 solutions.
For $a = -5$ we get $b+c = 5$ . $b$ can take values from -4 up to 2, hence 7 solutions.
For $a = -4$ we get $b+c = 4$ . $b$ can take values from -3 up to 1, hence 5 solutions.
For $a = -3$ we get $b+c = 3$ . $b$ can take values from -2 up to 1, hence 4 solutions.
For $a = -2$ we get $b+c = 2$ . $b$ can take values from -1 up to 0, hence 2 solutions.
For $a = -1$ we get $b+c = 1$ . $b$ can take values 0, hence 1 solutions.

Hence total of 14+13+11+10+8+7+5+4+2+1 = 75.

[Latex edits. Edits for clarity - Calvin]

We note that $a(b-c)^{3}+b(c-a)^{3}+c(a-b)^{3}=(a-b)(b-c)(c-a)(a+b+c)$

Since $a<b<c$ , the product $(a-b)(b-c)(c-a)$ is non-zero, and we must thus have $(a+b+c)=0$ , that is, $c=-a-b$ . The condition $(a+b+c)=0$ also implies that $a<0$ .

Now since $c>b$ , $-a-b>b$ , or $-a>2b$ . Then $\frac{-a}{2}>b$ .

Combining all the conditions, we find that each ordered pair $(a,b)$ satisfying the following inequalities $-10≤a<0$ $a<b<\frac{-a}{2}$ gives a unique solution to the original equation.

As $a$ ranges from $-10$ to $-1$ , the total number of solutions is $14+13+11+10+8+7+5+4+2+1=75$ .

NB: The factorization of the function can be motivated when one tries small values and realizes that (if one ignored the conditions on $a, b, c$ ) the equation would be satisfied by $a=b$ . By symmetry $b=c$ and $c=a$ also satisfy the equation. Thus, the product $(a-b)(b-c)(c-a)$ must be a factor of the given function and we can then obtain the remainder of the factorization by division.

Let f(a,b,c)=a(b−c)^3+b(c−a)^3+c(a−b)^3=a^3(c-b)+b^3(a-c)+c^3(b-a)

Note that f(a,b,c)=f(b,c,a) Thus, f(a,b,c) is cyclic.

f(a,a,c)=0, so (a-b) is a factor f(a,b,b)=0, so (b-c) is a factor f(c,b,c)=0, so (c-a) is a factor

deg(a-b)(b-c)(c-a)=3, but degf(a,b,c)=4, so we say f(a,b,c)=(a-b)(b-c)(c-a)(K(a+b+c)) for some constant K

To solve for K, we pick, (a,b,c)=(2,1,0)

a^3(c-b)+b^3(a-c)+c^3(b-a)=(a-b)(b-c)(c-a)(K(a+b+c))

2^3(0-1)+1^3(2-0)+0^3(1-2)=(2-1)(1-0)(0-2)(K(2+1+0)) -6=-6K K=1

Thus, f(a,b,c)=(a-b)(b-c)(c-a)(a+b+c)=0

If a-b=0 a=b, which cannot happen

If b-c=0 b=c, which cannot happen

If c-a=0 c=a, which cannot happen

This, a+b+c=0

We set a=-10 b+c=10, and b can be any value from -9 to 4, since it is less than c, but greater than a, so there are 14 cases. Similarly, a=-9 13 cases a=-8 11 cases a=-7 10 cases a=-6 8 cases a=-5 7 cases a=-4 5 cases a=-3 4 cases a=-2 2 cases a=-1 1 case if a>=0, then b,c>0 from the given condition, which makes it impossible for a+b+c=0

Therefore, the number of triplets are 14+13+11+10+8+7+5+4+2+1=75

Observe that if $f(a,b,c)$ is the multivariable polynomial given, then clearly $f(a,a,c) = f(a,b,b) = f(c,b,c) = 0$ . Then, this means the polynomial is divisible by $(a-b)(b-c)(c-a)$ . However, clearly this doesn't equal the given polynomial, and by simplifying things out, we find that $f(a,b,c) = (a-b)(b-c)(c-a)(a+b+c)$ . Then, since $a \ne b \ne c$ , we must have $a+b+c = 0$ . Then, considering the cases as $a$ ranges from $-10$ to $-1$ , we find that there are $14+13+11+10+8+7+5+4+2+1 = \boxed{75}$ solutions.

By simplifying the equation, the problem becomes $(a - b)(b - c)(c - a)(a + b + c) = 0$ . From $a \neq b$ , $b \neq c$ , and $c \neq a$ , we have $a + b + c = 0$ becomes the only way for us to gain the satisfying condition.

$a$ should ever be negative, else it makes $a + b + c > 0$ . Therefore, $-10 \leq a < 0$ . $c = - a - b$ for the sake of $a + b + c = 0$ . Finally, $b < - \frac {a}{2}$ , else it makes $c = b$ at $b = - \frac {a}{2}$ .

Ultimately, we have $-10 \leq a < 0$ , $a < b < - \frac {a}{2}$ , and $c = - a - b$ ; which therefore we could conclude that there is 75 integer triplets as the solution!

Since $a(b-c)^3+b(c-a)^3+c(a-b)^3=(b-a)(c-a)(c-b)(a+b+c)$ , $(b-a)(c-a)(c-b)(a+b+c)=0$ . From the problem, $a<b<c$ .

In other words, $a\neq{b}, b\neq{c}$ and $c\neq{a}$ . $b-a\neq0, c-a\neq0$ and $c-b\neq0.$ Therefore, $a+b+c=0$ . If $a\geq0, c>b>a\geq0$ and contradicts $a+b+c=0$ . Therefore, $-10\leq{a}\leq-1$ . When $a=-10$ , $b+c=10$ , since $-10=a<b<c$ and $b<5$ , $-9\leq{b}\leq4$ , and since a unique pair of $a$ and $b$ determines a unique value of $c$ , there are 14 solutions. Similarly, when $a=-9,-8,-7,-6,-5,-4,-3,-2,-1$ , there are 13,11,10,8,7,5,4,2,1 solutions respectively. Therefore, there are a total of 1+2+4+5+7+8+10+11+13+14=75 solutions.

Note that by substituting $a=b$ into the equation, we get $a(a-c)^3+a(c-a)^3+c(a-a)^3=a(a-c)^3-a(a-c)^3+c(0)^3=0$ so $a-b$ is a factor of the left hand side of the given equation. Similarly, $b-c$ and $c-a$ are also factors.

We can now proceed to factorize the given equation and by expanding and factorizing we get $(b-c)(c-a)(a-b)(a+b+c)=0$ . Since $a \neq b \neq c$ , $(b-c),(c-a),(a-b) \neq 0$ so it must be the case that $a+b+c=0$ .

$b<c=(-a-b) \Rightarrow 2b < -a \Rightarrow a < b < -\frac{a}{2}$ . We can thus deduce that $a < 0$ and $(a+1) \leq b \leq \lfloor \frac{-a-1}{2} \rfloor$ and we can consider each possible value of $a \in [-10,-1]$

$a=-10 \Rightarrow -9 \leq b \leq 4$ which contributes $4-(-9)+1=14$ triples

$a=-9 \Rightarrow -8 \leq b \leq 4$ which contributes $4-(-8)+1=13$ triples

$a=-8 \Rightarrow -7 \leq b \leq 3$ which contributes $3-(-7)+1=11$ triples

$a=-7 \Rightarrow -6 \leq b \leq 3$ which contributes $3-(-6)+1=10$ triples

$a=-6 \Rightarrow -5 \leq b \leq 2$ which contributes $2-(-5)+1=8$ triples

$a=-5 \Rightarrow -4 \leq b \leq 2$ which contributes $2-(-4)+1=7$ triples

$a=-4 \Rightarrow -3 \leq b \leq 1$ which contributes $1-(-3)+1=5$ triples

$a=-3 \Rightarrow -2 \leq b \leq 1$ which contributes $1-(-2)+1=4$ triples

$a=-2 \Rightarrow -1 \leq b \leq 0$ which contributes $0-(-1)+1=2$ triples

$a=-1 \Rightarrow 0 \leq b \leq 0$ which contributes $0-0+1=1$ triple

Therefore, there are a total of $14+13+11+10+8+7+5+4+2+1=75$ triples

So start from this, $b(c-a)^3=a(c-b)^3+c(b-a)^3$ $b(c-b+b-a)^3=a(c-b)^3+c(b-a)^3$ $b(c-b)^3+3b(c-b)^2(b-a)+3b(c-b)(b-a)^2+b(b-a)^3=a(c-b)^3+c(b-a)^3$ $(b-a)(c-b)^3+3b(c-b)^2(b-a)+3b(c-b)(b-a)^2+(b-c)(b-a)^3=0$ Since $b\ne a$ and $c\ne b$ , then $(c-b)^2+3b(c-b)+3b(b-a)-(b-a)^2=0$ $3b(c-a)+(c-2b+a)(c-a)=0$ Again, since $a\ne c$ , we have $3b+c-2b+a=0$ $a+b+c=0$ From $a+b+c=0$ , and $a<b<c$ , we can conclude that $a$ must be negative. So let $a=-x$ , then we have $b+c=x$ . If $x$ is even, then it must be $1-x\le b\le \dfrac{x}{2}-1$ , and the possible number for $b$ is $\dfrac{x}{2}-1-(1-x)+1=x+\dfrac{x}{2}-1$ . Else if $x$ is odd, then it must be $1-x\le b\le \dfrac{x-1}{2}$ , and the possible number for $b$ is $\dfrac{x-1}{2}-(1-x)+1=\dfrac{x-1}{2}+x$ . We know that $a$ is negative, so $-10\le a\le -1$ , equivalent with $1\le x\le 10$ . So total of the possible number for $b$ is $14+13+11+10+8+7+5+4+2+1=75$ Which means, $75$ is the total number of triples of integers that satisfied the condition.

a(b−c)^3+b(c−a)^3+c(a−b)^3= (a b^3 - 3 a b^2 c +3 a b c^2 -a c^3) + (b c^3 -3 b c^2 a +3 b c a^2 - b a^3) +

(c a^3 -3 c a^2 b +3 c a b^2 -c b^3)

a b^3 + b c^3 + c a^3 - a c^3 - b a^3 -c b^3 - 3 a b^2 c + 3 c a b^2 + 3 a b c^2 - 3 b c^2 a

+3 b c a^2 -3 c a^2 b

a b^3 + b c^3 + c a^3 - a c^3 - b a^3 -c b^3

a b (b^2-a^2) +a c (a^2-c^2) +b c (c^2-b^2) I haven't been able to find how to prove it but it appears from excel evaluations of three different forms of the expression, when the sum of a,b & c is zero then the expression also evaluates to zero a=-10 14 a=-9 13 a=-8 11 a= -7 10 a=-6 8 a=-5 7 a=-4 5 a=-3 4 a-2 2 a=-1 1 There are 75 triples of integers that meet the criteria

Let $f(a,b,c)$ denote the expression on the left-hand side of the given equation. Notice that if either $a=b$ or $b=c$ or $c=a$ , then $f(a,b,c)=0.$ This implies $(a-b)(b-c)(c-a)$ is a factor of $f(a,b,c).$ Since $f(a,b,c)$ is a polynomial of degree 4 and cyclic in $a,b,c$ , it follows that $f(a,b,c) = k(a+b+c)(a-b)(b-c)(c-a)$ for some constant $k.$ To determine $k$ , we simply let $(a,b,c) =(0,1,2)$ to obtain $6 = 6k$ or $k=1.$ Since it is required that $a<b<c$ , the given equation is equivalent to $a+b+c = 0.$ With restrictions that $a,b,c$ are integers with $-10 \le a < b < c$ , we can describe the solution set of this equation by $\bigcup_{k=-10}^{-1}\{(k,j,-k-j)\,:\, j=k+1,\dots,\lfloor -k/2\rfloor -1\}.$ The number of triples $(a,b,c)$ for $k=-10$ is the number of integers in $[-9,4]$ which is $14.$ Proceed in the same manner with $k=-9,-8,\dots,-1,$ we have the total number of triples $(a,b,c)$ with the given condition to be $14+13+11+10+8+7+5+4+2+1 = \boxed{75}.$

Since $a(b-c)^3+b(c-a)^3+c(a-b)^3=(b-a)(c-a)(c-b)(a+b+c)$ , $(b-a)(c-a)(c-b)(a+b+c)=0$ . From the problem, $a\neq{b},b\neq{c},c\neq{a}$ , this implies that $a+b+c=0$ .

If $a\geq0$ , $b,c>a\geq0$ and contradicts $a+b+c=0$ . Therefore, $-10\leq{a}\leq-1$ . When $a=-10, b+c=10$ , since $-10=a<b<c,-9\leq{b}\leq{4}$ and there are 14 solutions. Similarly, when $a=-9,-8,-7,-6,-5,-4,-3,-2,-1$ , there are $13,11,10,8,7,5,4,2,1$ solutions respectively. Therefore, there are a total of 75 solutions.

$a(b-c)^3 + b(c-a)^3 + c(a-b)^3 = 0$ $\Rightarrow (a-b)(b-c)(c-a)(a+b+c) = 0$

As, $a<b<c$ ,

$(a+b+c) = 0$

When $a=-10, b+c=10$ ; has 14 solutions.

When $a=-9, b+c=9$ ; has 13 solutions.

When $a=-8, b+c=8$ ; has 11 solutions.

When $a=-7, b+c=7$ ; has 10 solutions.

When $a=-6, b+c=6$ ; has 8 solutions.

When $a=-5, b+c=5$ ; has 7 solutions.

When $a=-4, b+c=4$ ; has 5 solutions.

When $a=-3, b+c=3$ ; has 4 solutions.

When $a=-2, b+c=2$ ; has 2 solutions.

When $a=-1, b+c=1$ ; has 1 solution.

Total = $1+2+4+5+7+8+10+11+13+14 = \boxed{75}$

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

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

Cause $a<b<c$ we have:

$a \neq b$

$b \neq c$

$c \neq a$

We have to solve $(a+b+c)=0$

That is $a = -(b+c)$

Counting cases, we have 14 cases if $a = -10$ , 13 if $a = -9$ , 11 if $a = -8$ , ... , 1 if $a = -1$ .

$14+13+11+10+8+7+5+4+2+1 = \boxed{75}$

The given equation eventually reduces to the form a+b+c=0.Put a= -10 and get 14 solutions of (b,c) so that a<b<c. Put a = -9 and get 13 solutions and go on till you have a=-1,b=0,c=1. So total no. of solutions is 75

If a is positive, then b-c, c-a, and a-b are all negative and therefore the sum cannot be 0. Given range of a as [-10, 0] I brute forced a solution in python.

reduce a(b-c)^3 + ... = 0 as --> a(b^2 + c^2) + b(a^2 + c^2) + c(b^2 + a^2) = 0 say eqn 1; consequently we know that for any a,b,c it follows that a(b-c) + b(c-a) + c(a-b) = 0 ; and (b-c) + (c-a) + (a-b) = 0; the above three are equations in terms of (a-b),(b-c),(c-a) .. for a non trivial solution to exist ..trivial is a=b=c=0; we must have that [a(b^2 + c^2) , b(a^2 + c^2),c(a^2 + b^2)] = k1[a,b,c]+k2[1,1,1]..

a(b^2 + c^2) = k1(a) + k2; b(c^2 + a^2) = k1(b) + k2; subtract both we have k1 = c^2 - ab .. similarly we have k1 = b^2 - ac ; k1 = a^2 - bc.

manipulate above 3 equations we get a + b + c = 0 ; since a<b<c and a>-11 .. substitute from a = -10 we get 14 cases .. and so on we end up with 75 solutions

a(b-c)^3 + b(c-a)^3 + c(a-b)^3 = 0

On expanding and simplifying we get;

ab^3 - cb^3 - ac^3 + ca^3 + bc^3 - ba^3 = 0

further,

-b^3(c-a) - ac(c^2-a^2) + b(c^3-a^3) = 0

-b^3 - ac^2 - a^2c + bc^2 + a^2b + abc = 0

b(c^2-b^2) + ac(b-c) + a^2(b-c) = 0

-b(b+c) + ac + a^2 = 0

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

a + b + c = 0

from this we can find out the possible no. of values of a and corresponding no. of b and c values as 75.