I want to factorize the polynomial $x^3+y^3+z^3-3xyz$. Making use of ptcouncil.netematica I discover that it amounts to $(x+y+z)(x^2+y^2+z^2-xy-yz-zx)$. But how can I factorize the by hand?


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Note the (can be conveniently seen with dominion of Sarrus)$$ \beginvmatrix x & y & z \\ z & x & y \\ y & z & x \\ \endvmatrix=x^3+y^3+z^3-3xyz$$

On the various other hand, the is equal to (if we add to the an initial row 2 various other rows)$$ \beginvmatrix x+y+z & x+y+z & x+y+z \\ z & x & y \\ y & z & x \\ \endvmatrix=(x+y+z)\beginvmatrix 1 & 1 & 1 \\ z & x & y \\ y & z & x \\ \endvmatrix=(x+y+z)(x^2+y^2+z^2-xy-xz-yz)$$ just as we wanted. The last equality complies with from the development of the determinant by very first row.

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answer Dec 27 "13 in ~ 15:17
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ElensilElensil
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Consider the polynomial $$(\lambda - x)(\lambda - y)(\lambda - z) = \lambda^3 - a\lambda^2+b\lambda-c\tag*1$$We know$$\begincasesa = x + y +z\\ b = xy + yz + xz \\ c = x y z\endcases$$ substitute $x, y, z$ for $\lambda$ in $(*1)$ and also sum, us get$$x^3 + y^3 + z^3 - a(x^2+y^2+z^2) + b(x+y+z) - 3c = 0$$This is equivalent to$$\beginalign x^3+y^3+z^3 - 3xyz= & x^3+y^3+z^3 - 3c\\= & a(x^2+y^2+z^2) - b(x+y+z)\\= & (x+y+z)(x^2+y^2+z^2 -xy - yz -zx)\endalign$$


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reply Oct 29 "13 at 11:16
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achille huiachille hui
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\beginalignx^3+y^3+z^3-3xyz\\&= x^3+y^3+3x^2y+3xy^2+z^3-3xyz-3x^2y-3xy^2\\&= (x+y)^3+z^3-3xy(x+y+z)\\&= (x+y+z)((x+y)^2+z^2-(x+y)z)-3xy(x+y+z)\\&= (x+y+z)(x^2+2xy+y^2+z^2-yz-xz-3xy)\\&= (x+y+z)(x^2+y^2+z^2-xy-yz-zx)\endalign


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edited Aug 23 at 21:40
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answered Oct 29 "13 at 10:23
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Use Newton"s identities:

$p_3=e_1 p_2 - e_2 p_1 + 3e_3$ and so $p_3-3e_3 =e_1 p_2 - e_2 p_1 = p_1(p_2-e_2)$ as required.

Here

$p_1= x+y+z = e_1$

$p_2= x^2+y^2+z^2$

$p_3= x^3+y^3+z^3$

$e_2 = xy + xz + yz$

$e_3 = xyz$


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edited Oct 29 "13 in ~ 11:31
achille hui
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answered Oct 29 "13 at 10:29
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A polynomial native $\ptcouncil.netbbQ$ is a polynomial from $\ptcouncil.netbbQ$, so it can be perceived as a polynomial in $z$ with coefficients from the integral domain $\ptcouncil.netbbQ$.$$p(z)=z^3-3xy \cdot z +x^3+y^3$$

So we can try our techniques to factor a polynomial of degree 3 end an integral domain:If it can be factored climate there is a element of level $1$, we call it $z-u(x,y)$ and $u(x,y)$ divides the continuous term that $p(z)$ i beg your pardon is $x^3+y^3$. The latter is have the right to be factored come $(x+y)(x^2-xy+y^2)$ We check each that the feasible values $(x+y), -(x+y), (x^2-xy+y^2), -(x^2-xy+y^2)$ because that $u(x,y)$ and also find that only $p(-x-y)=0$. Therefore $z-(-x-y)$ is a factor.

Note:

One deserve to use Kronecker"s technique

to mitigate the administrate of a polynomial that $\ptcouncil.netbbQ$ to factoring polynomials in $\ptcouncil.netbbQ$, to mitigate the administrate of polynomial of $\ptcouncil.netbbQ$ to factoring polynomials in $\ptcouncil.netbbQ$to mitigate the administer of polynomial of $\ptcouncil.netbbQ$ to factoring numbers in $\ptcouncil.netbbZ$

This factoring is possible in a finite number of steps but the variety of steps may end up being to high for valuable purpose.

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An integral domain is a commutative ring v $1$, where the adhering to holds:$$a \ne 0 \land b \ne 0 \implies abdominal muscle \ne 0$$For polynomials $f$, $g$, $h$ $\in I$ this guarantees:$$f=g \cdot h \implies \textdegree(f)=\textdegree(g) + \textdegree(h) \tag1$$compare this come $\ptcouncil.netbbZ_4$ which is no integral domain and also $(2z^2+1)^2 \equiv 1$ and also so $(2z^n+1) \mid 1$. Therefore the polynomial $1$ of degree $0$ has actually infinitely countless divisor.If $I$ is one integral domain $(1)$ guarantees that $z^3+az^2+bz+c \in I$ has a direct factor and therefore zero in $I$ if that is not irreduzible.