Slightly expanding, we have that $\frac{(a-b)\left(a^{2}+a b+b^{2}\right)}{(a-b)(a-b)(a-b)}=\frac{73}{3}$. Canceling the $(a-b)$, cross multiplying, and simplifying, we obtain that $0=70 a^{2}-149 a b+70 b^{2}$. Dividing everything by $b^{2}$, we get that $$ 0=70\left(\frac{a}{b}\right)^{2}-149\left(\frac{a}{b}\right)+70 \text {. } $$ Applying the quadratic formula and following the restriction that $a\textgreater b\textgreater0$ $$ \frac{a}{b}=\frac{10}{7} \text {. } $$ Hence, $7 a=10 b$. Since they are relatively prime, $a=10, b=7$. $$ 10-7= 3 \text {. } $$ 
Let $a$ and $b$ be relatively prime positive integers with $a\textgreater b\textgreater0$ and $\frac{a^{3}-b^{3}}{(a-b)^{3}}=\frac{73}{3}$. What is $a-b$? 
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