Question

Find the zero of the polynomial f(x) = x³-3 and verify the relation between their zeros and coefficient of polynomial.

Answer 1

$$\begin{gathered} Rootsof{x}^3-1=0are1,\omega ,{\omega }^{2}where \\ 1+\omega +{\omega }^2=0and{\omega }^3=1 \\ Sorootsof{x}^3-3=0are\sqrt[3]{3},\sqrt[3]{3}\omega ,\sqrt[3]{3}{\omega }^2. \\ Verification: \\ Comparing{x}^3-3=0witha{x}^3+b{x}^2+cx+d=0weget \\ a=1;b=0;c=0andd=-3; \\ Sumofroots=\frac{-b}{a} \\ \sqrt[3]{3}+\sqrt[3]{3}\omega +\sqrt[3]{3}{\omega }^2=\frac{0}{1} \\ \sqrt[3]{3}(1+\omega +{\omega }^2)=0 \\ \sqrt[3]{3}\left(0\right)=0 \\ 0=0,soverified. \\ Sumofproductsofrootstaken2atatime=\frac{c}{a} \\ \sqrt[3]{3}*\sqrt[3]{3}\omega +\sqrt[3]{3}\omega *\sqrt[3]{3}{\omega }^2+\sqrt[3]{3}{\omega }^2*\sqrt[3]{3}=\frac{0}{1} \\ {\left(\sqrt[3]{3}\right)}^2\omega +{\left(\sqrt[3]{3}\right)}^2{\omega }^3+{\left(\sqrt[3]{3}\right)}^2{\omega }^2=0 \\ {\left(\sqrt[3]{3}\right)}^2\omega +{\left(\sqrt[3]{3}\right)}^2*1+{\left(\sqrt[3]{3}\right)}^2{\omega }^2=0 \\ {\left(\sqrt[3]{3}\right)}^2(\omega +1+{\omega }^2)=0 \\ {\left(\sqrt[3]{3}\right)}^2\left(0\right)=0 \\ 0=0,soverified \\ Productofallroots=\frac{-d}{a} \\ \sqrt[3]{3}*\sqrt[3]{3}\omega *\sqrt[3]{3}{\omega }^2=\frac{-(-3)}{1} \\ {\left(\sqrt[3]{3}\right)}^3*{\omega }^3=3 \\ 3\left(1\right)=3 \\ 3=3,soverified \end{gathered}$$