If $(a,b)=1$, then $ \gcd(a^2+b^2,a^3+b^3)\mid (a-b)$ 
If $(a,b)=1$, then $\gcd(a^2+b^2,a^3+b^3)\mid (a-b)$

The only way that can help is to find some common factor of $a^2+b^2$ and $a^3+b^3$. 
That does not seem obvious enough, so will directly try to divide $a^3+b^3$ by $a^2+b^2$.
This leads to nowhere too.
It seems that need to use the fact that $a,b$ are relatively prime, but am unable to use that.
 Let for some suitable integers $x, y$, have $ax +by =1$. 
 A: Say $$d=\gcd(a^2+b^2,a^3+b^3)$$ then $d\mid (a+b)(a^2+b^2-ab)$ and $d\mid a^2+b^2$ so, we get: $$d\mid ab(a+b) = (a+b)(a^2+b^2)-  (a+b)(a^2+b^2-ab)$$
Now suppose there is prime $p$ such that $p\mid d$ and $p\mid a$.  Then $p\mid a^2$ and so $p\mid b^2 = (a^2+b^2)-a^2$. A contradiction, since $a,b$ are relatively prime. So $a,d$ are relatively prime (and $b,d$ also) and $$d\mid a+b\Longrightarrow d\mid a(a+b)=a^2+ab$$
so $$ d\mid (a^2+ab)- (a^2+b^2)= b(a-b)$$
By Euclid lemma we have $d\mid a-b$.
A: Note that
$$
\left(a^3+b^3\right)-b\left(a^2+b^2\right)=(a-b)a^2
$$
and
$$
a\left(a^2+b^2\right)-\left(a^3+b^3\right)=(a-b)b^2
$$
and if
$$
ax+by=1
$$
then
$$
\begin{align}
1
&=(ax+by)^3\\
&=a^2\left(ax^3+3x^2by\right)+b^2\left(3axy^2+by^3\right)
\end{align}
$$
Thus,
$$
\begin{align}
a-b
&=(a-b)\left(\color{#C00}{a^2}\left(ax^3+3x^2by\right)+\color{#090}{b^2}\left(3axy^2+by^3\right)\right)\\
&=\color{#C00}{\left(\left(a^3+b^3\right)-b\left(a^2+b^2\right)\right)}\left(ax^3+3x^2by\right)\\
&+\color{#090}{\left(a\left(a^2+b^2\right)-\left(a^3+b^3\right)\right)}\left(3axy^2+by^3\right)\\
&=\boldsymbol{\left(a^3+b^3\right)}\left(ax^3+3x^2by-3axy^2-by^3\right)\\
&+\boldsymbol{\left(a^2+b^2\right)}\left(3a^2xy^2+aby^3-abx^3-3x^2b^2y\right)
\end{align}
$$
which means that
$$
\left.\left(a^2+b^2,a^3+b^3\right)\,\middle|\,a-b\right.
$$
