2x^3+3x-1=0
a=2 , b=3 , c=-1
Wzory Viete’a
\alpha + \beta =-\frac{b}{2a}=-\frac{3}{2}
\alpha\beta=-\frac{c}{a}=-\frac{1}{2}
Wyprowadzam \alpha+\beta^3 ze wzoru skróconego mnożenia
(\alpha+\beta)^3=\alpha^3+3\alpha^2\beta+3\alpha\beta^2+\beta^3
(\alpha+\beta)^3=\alpha^3+\beta^3+3\alpha^2\beta+3\alpha\beta^2
(\alpha+\beta)^3=\alpha^3+\beta^3+3\alpha\beta(\alpha+\beta)
(\alpha+\beta)^3-3\alpha\beta(\alpha+\beta)=\alpha^3+\beta^3
Podstawiam wartości \alpha+\beta i \alpha\beta
{\alpha^3+\beta^3=(-\frac{3}{2})^3-3\cdot (-\frac{1}{2})\cdot (-\frac{3}{2})=-\frac{27}{8}-3\cdot \frac{3}{4}=-\frac{27}{8}-\frac{9}{4}=-\frac{27}{8}-\frac{18}{8}=-\frac{45}{8}}=-5\frac{5}{8}
\alpha^3+\beta^3=-\frac{45}{8}=-5\frac{5}{8}