\[\begin{array}{l} {\text{Пусть }}a,b,c{\text{ - корни многочлена }}{x^3} - {x^2} - 2x + 1. \hfill \\ {\text{Докажите тождества:}} \hfill \\ {\text{а) }}\sqrt[3]{a} + \sqrt[3]{b} + \sqrt[3]{c} = \sqrt[3]{{3\sqrt[3]{7} - 5}}; \hfill \\ {\text{б) }}\sqrt[3]{{a + b}} + \sqrt[3]{{a + c}} + \sqrt[3]{{b + c}} = \sqrt[3]{{3\sqrt[3]{7} - 4}}; \hfill \\ {\text{в) }}\sqrt[3]{{a{b^2}}} + \sqrt[3]{{{a^2}c}} + \sqrt[3]{{b{c^2}}} = \sqrt[3]{7}; \hfill \\ {\text{г) }}\sqrt[3]{{\frac{{{a^2}}}{{{b^2}}}}} + \sqrt[3]{{\frac{{{b^2}}}{{{c^2}}}}} + \sqrt[3]{{\frac{{{c^2}}}{{{a^2}}}}} = \sqrt[3]{{49}}; \hfill \\ {\text{д) }}\sqrt[3]{{\frac{{{a^4}}}{{{b^2}{c^2}}}}} + \sqrt[3]{{\frac{{{b^4}}}{{{a^2}{c^2}}}}} + \sqrt[3]{{\frac{{{c^4}}}{{{a^2}{b^2}}}}} = 5; \hfill \\ {\text{е) }}\frac{1}{{\sqrt[3]{{a + 1}}}} + \frac{1}{{\sqrt[3]{{b + 1}}}} + \frac{1}{{\sqrt[3]{{c + 1}}}} = 0. \hfill \\ \end{array}\]