Giải thích các bước giải:
Định lí Vi-et: \(x_{1}+x_{2}=\frac{5}{3}\)
\(x_{1}.x_{2}=\frac{7}{3}\)
a. \(\frac{1}{x_{1}^{3}}+\frac{1}{x_{2}^{3}}=\frac{x_{1}^{3}+x_{2}^{3}}{(x_{1}.x_{2})^{3}}=\frac{(x_{1}+x_{2})(x_{1}^{2}-x_{1}x_{2}+x^{2})}{(x_{1}.x_{2})^{3}}=\frac{(x_{1}+x_{2})[(x_{1}+x_{2})^{2}-3x_{1}x_{2}]}{(x_{1}.x_{2})^{3}}=\frac{\frac{5}{3}[(\frac{5}{3})^{2}-3.\frac{7}{3}]}{(\frac{7}{3})^{2}}=\frac{-190}{147}\)
b. \(x_{1}^{4}-x_{2}^{4}=(x_{1}^{2}-x_{2}^{2})(x_{1}^{2}+x_{2}^{2}=[(x_{1}+x_{2})(x_{1}-x_{2})][(x_{1}+x_{2})^{2}-2x_{1}x_{2}]=[(x_{1}+x_{2})(\frac{x_{1}^{3}-x_{2}^{3}}{x_{1}^{2}+x_{1}x_{2}+x_{2}^{2}})][(x_{1}+x_{2})^{2}-2x_{1}x_{2}]=[(x_{1}+x_{2})((x_{1}+x_{2})[(x_{1}+x_{2})^{2}-3x_{1}x_{2}])][(x_{1}+x_{2})^{2}-2x_{1}x_{2}]=[\frac{5}{3}.\frac{5}{3}[(\frac{5}{3})^{2}-3.\frac{7}{3}][((\frac{5}{3})^{2}-2.\frac{7}{3}]=\)
