An algebra problem by John Paolo Bayaborda

ab = 1369 $\sqrt{2}$ ; $\frac{a}{b}$ = $\frac{\sqrt{2}}{2}$

by transposition, a = $\frac{b\sqrt{2}}{2}$

by substitution, $\frac{b^2 \sqrt{2}}{2}$ = 1369 $\sqrt{2}$

therefore, b = 37 $\sqrt{2}$ and a = 37.

$x^3$ ( $\frac{a}{x^3}+\frac{1}{x^2}+\frac{1}{x}+1=0$ ), so, it will be equal to a + x + $x^2$ + $x^3$ = 0

therefore, x + $x^2$ + $x^3$ = -a

so, x + $x^2$ + $x^3$ + b = -a + b

and the answer is 37 $\sqrt{2}$ - 37 = $\boxed{15.33}$