I want to factorize the polynomial \$x^3+y^3+z^3-3xyz\$. Using orsini-gotha.comematica I find that it equals \$(x+y+z)(x^2+y^2+z^2-xy-yz-zx)\$. But how can I factorize it by hand?

eginalignx^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

Consider the polynomial \$\$(lambda - x)(lambda - y)(lambda - z) = lambda^3 - alambda^2+blambda-c ag*1\$\$We know\$\$egincasesa = x + y +z\ b = xy + yz + xz \ c = x y zendcases\$\$ Substitute \$x, y, z\$ for \$lambda\$ in \$(*1)\$ và sum, we get\$\$x^3 + y^3 + z^3 - a(x^2+y^2+z^2) + b(x+y+z) - 3c = 0\$\$This is equivalent to\$\$eginalign x^3+y^3+z^3 - 3xyz= và 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\$\$

Note that (can be easily seen with rule of Sarrus)\$\$ eginvmatrix x & y và z \ z và x & y \ y & z & x \ endvmatrix=x^3+y^3+z^3-3xyz\$\$

On the other hand, it is equal to lớn (if we showroom to the first row 2 other rows)\$\$ eginvmatrix x+y+z và x+y+z và x+y+z \ z & x & y \ y và z & x \ endvmatrix=(x+y+z)eginvmatrix 1 và 1 & 1 \ z & x và y \ y & z & x \ endvmatrix=(x+y+z)(x^2+y^2+z^2-xy-xz-yz)\$\$ just as we wanted. The last equality follows from the expansion of the determinant by first row.

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answered Dec 27, 2013 at 15:17

ElensilElensil
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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, 2013 at 11:31
achille hui
answered Oct 29, 2013 at 10:29
lhflhf
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A polynomial from \$orsini-gotha.combbQ\$ is a polynomial from \$orsini-gotha.combbQ\$, so it can be viewed as a polynomial in \$z\$ with coefficients from the integral domain name \$orsini-gotha.combbQ\$.\$\$p(z)=z^3-3xy cdot z +x^3+y^3\$\$

So we can try our methods khổng lồ factor a polynomial of degree 3 over an integral domain:If it can be factored then there is a factor of degree \$1\$, we call it \$z-u(x,y)\$ và \$u(x,y)\$ divides the constant term of \$p(z)\$ which is \$x^3+y^3\$. The latter is can be factored to \$(x+y)(x^2-xy+y^2)\$ We kiểm tra each of the possible values \$(x+y), -(x+y), (x^2-xy+y^2), -(x^2-xy+y^2)\$ for \$u(x,y)\$ & find that only \$p(-x-y)=0\$. So \$z-(-x-y)\$ is a factor.

Note:

One can use Kronecker"s method

to reduce the factorization of a polynomial of \$orsini-gotha.combbQ\$ to lớn factoring polynomials in \$orsini-gotha.combbQ\$, to reduce the factorization of polynomial of \$orsini-gotha.combbQ\$ lớn factoring polynomials in \$orsini-gotha.combbQ\$to reduce the factorization of polynomial of \$orsini-gotha.combbQ\$ to lớn factoring numbers in \$orsini-gotha.combbZ\$

This factoring is possible in a finite number of steps but the number of steps may become khổng lồ high for practical purpose.

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An integral tên miền is a commutative ring with \$1\$, where the following holds:\$\$a e 0 land b e 0 implies ab e 0\$\$For polynomials \$f\$, \$g\$, \$h\$ \$in I\$ this guarantees:\$\$f=g cdot h implies extdegree(f)= extdegree(g) + extdegree(h) ag1\$\$compare this to lớn \$orsini-gotha.combbZ_4\$ which is no integral domain and \$(2z^2+1)^2 equiv 1\$ and so \$(2z^n+1) mid 1\$. So the polynomial \$1\$ of degree \$0\$ has infinitely many divisor.If \$I\$ is an integral tên miền \$(1)\$ guarantees that \$z^3+az^2+bz+c in I\$ has a linear factor và therefore zero in \$I\$ if it is not irreduzible.