Question

Vreo formula pentru x1^3+x2^3+x3^3 polinoame ​

Answer 1

$(x+y+z)^{3}=x^{3}+y^{3}+z^{3}+6xyz+3x^{2}y+3x^{2}z+3y^{2}x+3y^{2}z+3z^{2}z+3z^{2}y\\\\(x+y+z)^{3}=x^{3}+y^{3}+z^{3}+(3x^{2}y+3y^{2}x+3xyz)+(3x^{2}z+3z^{2}x+3xyz)+(3z^{2}y+3y^{2}z+3xyz)-3xyz\\\\(x+y+z)^{3}=x^{3}+y^{3}+z^{3}+3xy(x+y+z)+3yz(x+y+z)+3xz(x+y+z)-3xyz\\(x+y+z)^{3}=x^{3}+y^{3}+z^{3}+3(x+y+z)(xy+yz+xz)-3xyz\\\\\\\\x^{3}+y^{3}+z^{3}=(x+y+z)^{3}-3(x+y+z)(xy+yz+xz)+3xyz$

x,y,z sunt radacinile ecuatiei de gr 3(doar inlocuiesti x,y,z, cu x1,x2,x3)

Polinom de grad 3:

$a_{3}X^{3}+a_{2}X^{2}+a_{1}X+a_{0}=0$

Relatiile lui Viete:

$x_{1}+x_{2}+x_{3}=-\frac{a_{2}}{a_{3}} \\x_{1}x_{2}+x_{2}x_{3}+x_{1}x_{3}=\frac{a_{1}}{a_{3}} \\x_{1}x_{2}x_{3}=-\frac{a_{0}}{a_{3}}$

Rezultat final:

$x_{1}^{3}+x_{2}^{3}+x_{3}^{3}=(-\frac{a_{2}}{a_{3}} )^{3}-3(\frac{-a_{2}}{a_{3}} )(\frac{a_{1}}{a_{3}} )+3(\frac{-a_{0}}{a_{3}} )$