Factorization of polynomial $(a-b)^4+(b-c)^4+(c-a)^4$

Question

Polynomial manifactoring of $(a-b)^4+(b-c)^4+(c-a)^4$ I've tried to factor all but didn't help. I'm in sevent grade and don't know many techniques So I wonder could someone help me.

Answer 1

Let $x=a-b$ and $y=b-c$ then $a-c=x+y$ so we have \begin{align}...&= x^4+y^4+(x+y)^4\\ &= 2x^4+4x^3y+6x^2y^2+4y^3x+2y^4\\ &= 2(x^4+2x^3y+3x^2y^2+2y^3x+y^4)\\ &= 2(x^2+xy+y^2)^2 \end{align}

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In general:

If $f(x,y)$ is 4 degree polynomial then you can try factor it by setting

$$f(x,y) = (ax^2+bxy+cy^2)(dx^2+exy+fy^2)$$ or $$f(x,y) = (ax+by)(cx^3+dx^2y+exy^2 +fy^3)$$ Now all you have to find are those coefficents.

Answer 2

We have \begin{align*} (a-b)^4+(b-c)^4+(c-a)^4 & = 2(a^2 - ab - ac + b^2 - bc + c^2)^2, \end{align*}

Answer 3

One method is based on a simple observation and a simple guess. Given the polynomial $\,P :=(a-b)^4+(b-c)^4+(c-a)^4\,$ to factor, notice that if $\,a,b,c\,$ are all increased or decreased by the same quantity, then the differences are unchanged. Thus, if $\,x=a-b,\; y=b-c\,$ then $$ P = x^4+y^4+(-x-y)^4 = 2(x^4+2x^3y+3x^2y^2+2xy^3+y^4). $$ If $\,y=1\,$ and $\,x\,$ is a small integer, then the polynomial factor is a square. For example, $\,x=y=1\,$ implies $\, P = 2\cdot3^2.\,$ A good guess is that it is a square of the form $\, P =2(ux^2+vxy+wy^2)^2\,$ for some integers $\,u,v,w\,$ and a good guess is that $\,u=v=w=1\,$ which is correct. Thus, $$ P = 2(x^2+xy+y^2)^2 = 2((a-b)^2+(a-b)(b-c)+(b-c)^2)^2. $$

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Another way is to find a systematic solution, but it requires some advanced techniques that you almost certainly do not know yet. It uses recursion relations and elementary symmetric polynomials which in this case are $$ e_1 := x+y+z,\quad e_2 := xy+yz+zx,\quad e_3 :=xyz. \tag{1} $$ Define the power symmetric polynomials $$ p_n := x^n + y^n + z^n. \tag{2} $$ The fundamental theorem of symmetric polynomials is that all such polynomials can be expressed using the elementary symmetric polynomials. The base case $\,p_0 = 3\,$ is immediate. Next is, $\, p_1 = x+y+z = e_1.\,$ An easy calculation gets $$ p_2 = x^2+y^2+z^2 = (x+y+z)^2-2(xy+yz+zx) = e_1^2-2e_2. \tag{3} $$ Note that $$ e_2 = (ab+bc+ca)-(a^2+b^2+c^2). \tag{4} $$ In general, the power symmetric polynomials satisfy a recursion $$ p_{n+3} = e_1 p_{n+2} - e_2 p_{n+1} + e_3 p_n. \tag{5} $$

In our particular case, $$ x = a-b,\quad y = b-c,\quad z = c-a \tag{6} $$ which implies using equation $(3)$ that $$ p_1 = e_1 = 0,\quad p_2 = -2e_2. \tag{7} $$ Use the recursion in equation $(5)$ to get $$ p_3 = 3e_3,\quad p_4 = 2e_2^2. \tag{8} $$ Translate this in terms of $\,a,b,c\,$ to get $$ p_3 = 3(a-b)(b-c)(c-a),\quad p_4 = 2e_2^2. \tag{9} $$