Given.

$\alpha ,\beta ,\gamma$ are the roots of the equation $x^3+x^2+x+r=0$

So, $\alpha +\beta +\gamma =-1 ....\left(1\right)$

And $\alpha \beta +\beta \gamma +\gamma \alpha =1 .......\left(2\right)$

And $\alpha \beta \gamma =-r ........\left(3\right)$

Now we know that,

${\alpha }^3+{\beta }^3+{\gamma }^3-3\alpha \beta \gamma =\left(\alpha +\beta +\gamma \right)\left({\alpha }^2+{\beta }^2+{\gamma }^2-\alpha \beta -\beta \gamma -\alpha \gamma \right)$

Now putting the value from all above equation in formula we get,

$\Rightarrow 5+3r=\left(-1\right)\left({\left(\alpha +\beta +\gamma \right)}^2-3\left(\alpha \beta +\beta \gamma +\alpha \gamma \right)\right)$ 

{as ${\alpha }^3+{\beta }^3+{\gamma }^3=5$}

$\Rightarrow 5+3r=\left(-1\right)\left(1-3\left(1\right)\right)$

$\Rightarrow 5+3r=2$

$\Rightarrow r=-1$