Suppos a a, b, c are the roots of the cubic ${\mathrm{x}}^3-{\mathrm{x}}^2-2=0$. Then the value of a³ + b³ + c³ is ________.

Given a + b + c = 1                    …….. (1)

ab + bc + ca = 0                        …….. (2)

abc = 2                                      ……… (3)

Now (a + b + c)² = 1

$\begin{array}{l}{\mathrm{a}}^2+{\mathrm{b}}^2+{\mathrm{c}}^2+2\sum \mathrm{ab}=1\\ \therefore {\mathrm{a}}^2+{\mathrm{b}}^2+{\mathrm{c}}^2=1\end{array}$

Now, ${\mathrm{a}}^3+{\mathrm{b}}^3+{\mathrm{c}}^3-3\mathrm{abc}=(\mathrm{a}+\mathrm{b}+\mathrm{c})\left[\sum {\mathrm{a}}^2-\sum \mathrm{ab}\right]$

                                        $=1(1-0)=1$

$\therefore {\mathrm{a}}^3+{\mathrm{b}}^3+{\mathrm{c}}^3=1+3\mathrm{abc}=1+3\times 2=7$