Algebra

a binomial

 

$a=5$   ;   $b=-11$   ;   $c=4$

 

pluggint hat info into    $_a x_b=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$   we get:

 

$_a x_b= \frac{-\left(-11\right)\pm \sqrt{\left(-11\right)^2-4 \times \:5 \times \:4}}{2 \times\:5}$

 

$    _a x_b=\frac{11 \pm \sqrt{41}}{10}  $

 

$ a=1.74  $    and $  b=0.46 $

 

now for our condition where we have to find $a^3 + b^3  \ \ \implies \ \   1.74^3 +0.46^3 \ \ \implies \ \ \   5.27 + 0.097  $

 

so the answer is $  \boxed { 5.367  } $