I want to lớn factorize the polynomial \$x^3+y^3+z^3-3xyz\$. Using xemlienminh360.netematica 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

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

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

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answered Dec 27 "13 at 15:17

ElensilElensil
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Consider the polynomial \$\$(lambdomain authority - x)(lambdomain authority - y)(lambdomain authority - 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)\$ & 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\= và a(x^2+y^2+z^2) - b(x+y+z)\= và (x+y+z)(x^2+y^2+z^2 -xy - yz -zx)endalign\$\$

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answered Oct 29 "13 at 11:16

achille huiachille hui
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Use Newton"s identities:

\$p_3=e_1 p_2 - e_2 p_1 + 3e_3\$ và 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 at 11:31
achille hui
answered Oct 29 "13 at 10:29
lhflhf
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A polynomial from \$xemlienminh360.netbbQ\$ is a polynomial from \$xemlienminh360.netbbQ\$, so it can be viewed as a polynomial in \$z\$ with coefficients from the integral tên miền \$xemlienminh360.netbbQ\$.\$\$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 Gọi 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)\$ và find that only \$p(-x-y)=0\$. So \$z-(-x-y)\$ is a factor.

Note:

One can use Kronecker"s method

khổng lồ reduce the factorization of a polynomial of \$xemlienminh360.netbbQ\$ to lớn factoring polynomials in \$xemlienminh360.netbbQ\$, khổng lồ reduce the factorization of polynomial of \$xemlienminh360.netbbQ\$ khổng lồ factoring polynomials in \$xemlienminh360.netbbQ\$lớn reduce the factorization of polynomial of \$xemlienminh360.netbbQ\$ to factoring numbers in \$xemlienminh360.netbbZ\$

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

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 lớn \$xemlienminh360.netbbZ_4\$ which is no integral domain & \$(2z^2+1)^2 equiv 1\$ và so \$(2z^n+1) mid 1\$. So the polynomial \$1\$ of degree \$0\$ has infinitely many divisor.If \$I\$ is an integral domain name \$(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.

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