$a^{3} + b^{3}=?$

$\begin{aligned} \frac{a}{b} + \frac{b}{a} = 1 \end{aligned}$ $\begin{aligned} a^2 +b^2 = ab \Rightarrow {\color{#3D99F6}(a^2-ab+b^2)}= 0\end{aligned}$

We have $\begin{aligned} a^3 + b^3 = ( a+b){\color{#3D99F6}(a^2-ab+b^2)} = (a+b)*0 \Rightarrow \boxed{0}\end{aligned}$

$\dfrac{b}{a}=x$

So

$\dfrac{1}{x}+x = 1$

$\implies 1+x^2-x=0$

$\implies (x+1)(x^2-x+1)=0$ [On one condition, such that x $\neq$ -1]

$\implies x^3+1=0$

$\implies b^3+a^3=0$

$\implies a^3+b^3=\boxed{0}$

$\begin{aligned} \frac ab + \frac ba & = 1 \\ \frac {a^2+b^2}{ab} & = 1 \\ \implies \color{#3D99F6} a^2+b^2 & \color{#3D99F6} = ab & \color{#3D99F6}...(1) \\ (a+b)^2 & = {\color{#3D99F6}a^2} + 2ab + {\color{#3D99F6}b^2} & \small \color{#3D99F6} (1): a^2+b^2 = ab \\ & = 3ab \\ \implies \color{#D61F06} a+b & \color{#D61F06} = \sqrt{3ab} & \color{#D61F06} ...(2) \end{aligned}$

Now, we have:

$\begin{aligned} (a+b)^3 & = (\sqrt{3ab})^3 \\ a^3 + 3a^2b + 3ab^2 + b^3 & = 3ab\sqrt{3ab} \\ a^3 + 3ab{\color{#D61F06}(a+b)} + b^3 & = 3ab\sqrt{3ab} & \small \color{#D61F06} (2): a+b = \sqrt{3ab} \\ a^3 + 3ab{\color{#D61F06}\sqrt{3ab}} + b^3 & = 3ab\sqrt{3ab} \\ \implies a^3 + b^3 & = \boxed{0} \end{aligned}$

All the solutions here are correct, but notice that a and b can't be simultaneously real , as you can see that $\frac{a}{b}+\frac{b}{a}\geqslant 2$ for all real $a,\ b$ .