Note that $a-1=\frac{x}{y}$ and $b-1=\frac{y}{x}$ are reciprocals. That is, $$(a-1)(b-1)=1 \Longrightarrow a b-a-b+1=1 \Longrightarrow a b=a+b$$ Let $t=a b=a+b$. Then we can write $$a^{2}+b^{2}=(a+b)^{2}-2 a b=t^{2}-2 t$$ so $t^{2}-2 t=15$, which factors as $(t-5)(t+3)=0$. Since $a, b>0$, we must have $t=5$. Then, we compute $$a^{3}+b^{3}=(a+b)^{3}-3 a b(a+b)=5^{3}-3 \cdot 5^{2}=50$$