Let $(k,m)=(ga,gb)$ where $g=\gcd(k,m)$ (and $a>b$). Now $k^3-m^3|km(k^2-m^2)\implies a^3-b^3|gab(a^2-b^2)\implies a^2+ab+b^2|gab(a+b)$. Suppose there exists a prime $p$ such that $p|a^2+ab+b^2$ and $p|ab(a+b)$. So $p$ divides one of $a,b,a+b$. If $p|a$ then $p|a^2+ab$ but also $a^2+ab+b^2$ i.e. $p|b^2$, a contradiction since $a,b$ are coprime. Same goes for the case $p|b$. If $p|a+b$ then $p|(a+b)^2=a^2+2ab+b^2$. Then $p|a^2+2ab+b^2-(a^2+ab+b^2)=ab$ so we are back to $p$ dividing one of $a,b$, which we have seen to be impossible. Hence $\gcd(a^2+ab+b^2,ab(a+b))=1$ so $a^2+ab+b^2|g$ - this implies $g\ge a^2+ab+b^2$. Now $a^2+ab+b^2>3ab$ by AM-GM ($a\not= b$) so that $(a^2+ab+b^2)(a-b)^3>3ab$ since $(a-b)$ is positive and at least one. But $g(a-b)^3\ge (a^2+ab+b^2)(a-b)^3>3ab\implies g^3(a-b)^3>3g^2ab$ i.e. $(k-m)^3>3km$.

If $k^3-m^3|km(k^2-m^2)$, then dividing by $k-m$ we have $k^2+km+m^2|km(k+m)=k^2m+km^2$. Trivially, $k^2+km+m^2|k(k^2+km+m^2)=k^3+k^2m+km^2$, which implies $k^2+km+m^2|k^3$. Similarly, $k^2+km+m^2|m^3$. Let $gcd(k,m)=d$. Since then $gcd(k^3,m^3)=d^3$, there exists a linear combination $Ak^3+Bm^3=d^3$. Hence $k^2+km+m^2|d^3$. But then $k^2+km+m^2|d^3\left(\frac kd-\frac md\right)^3=(k-m)^3$. Since these are positive numbers, this means that $(k-m)^3$ cannot be smaller than $k^2+km+m^2$, therefore $(k-m)^3\ge k^2+km+m^2>3km$.

$gcd(k,m)=g,k=ga,m=gb,gcd(a,b)=1$ $$a^2+ab+b^2|gab(a+b)$$$$a^2+ab+b^2|g \implies g >3ab$$$$(k-m)^3 \ge g^3 > 3g^2ab$$

AK1024 wrote: Let $(k,m)=(ga,gb)$ where $g=\gcd(k,m)$ (and $a>b$). Now $k^3-m^3|km(k^2-m^2)\implies a^3-b^3|gab(a^2-b^2)\implies a^2+ab+b^2|gab(a+b)$. Suppose there exists a prime $p$ such that $p|a^2+ab+b^2$ and $p|ab(a+b)$. So $p$ divides one of $a,b,a+b$. If $p|a$ then $p|a^2+ab$ but also $a^2+ab+b^2$ i.e. $p|b^2$, a contradiction since $a,b$ are coprime. Same goes for the case $p|b$. If $p|a+b$ then $p|(a+b)^2=a^2+2ab+b^2$. Then $p|a^2+2ab+b^2-(a^2+ab+b^2)=ab$ so we are back to $p$ dividing one of $a,b$, which we have seen to be impossible. Hence $\gcd(a^2+ab+b^2,ab(a+b))=1$ so $a^2+ab+b^2|g$ - this implies $g\ge a^2+ab+b^2$. Now $a^2+ab+b^2>3ab$ by AM-GM ($a\not= b$) so that $(a^2+ab+b^2)(a-b)^3>3ab$ since $(a-b)$ is positive and at least one. But $g(a-b)^3\ge (a^2+ab+b^2)(a-b)^3>3ab\implies g^3(a-b)^3>3g^2ab$ i.e. $(k-m)^3>3km$. a, b can be composite

Observe that $k^3-m^3 \mid km(k^2-m^2) \implies k^2+km+m^2 \mid km(k+m)=k^2m+km^2$. The key step is the following: $$ k^2+km+m^2 \mid m(k^2+km+m^2) - k^2m-km^2 = m^3 $$Suppose that $\gcd(k,m) =d$ with $m=dm_1$, $km=k_1d$ meaning that $\gcd(k_1, m_1) =1$. This reveals that: $$ k^2+km+m^2 \mid m^3 \implies k_1^2+k_1m_1+m_1^2 \mid dm_1^3 $$Now just note that $\gcd (m_1, k_1^2+k_1m_1+m_1^2)=1$, because if not then there exists prime $p$ which divides both $m_1$ and $ k_1^2+k_1m_1+m_1^2$ meaning that $p \mid k_1$. This contradicts with $\gcd(k_1, m_1) =1$. Therefore we deduce that: $$ k_1^2+k_1m_1+m_1^2 \mid d \implies d \ge k_1^2+k_1m_1+m_1^2 \implies d^3 \ge d^2(k_1^2+k_1m_1+m_1^2)=k^2+km+m^2 \ge 3km $$where the lat inequality follows from AM-GM. This finishes the problem because $$ (k-m)^3 = d^3(k_1-m_1)^3 >d^3 \ge 3km $$as needed.

Let $d=\gcd (k,m)\Rightarrow \left\{ \begin{array}{l} k = dx\\ m = dy \end{array} \right.(x,y\in \mathbb{Z}^+,x > y,\gcd (x,y) = 1)$ We have $km(k^2-m^2)\vdots k^3-m^3\Rightarrow km(k+m)\vdots k^2+km+m^2\Rightarrow d^3xy(x+y)\vdots d^2(x^2+xy+y^2)$ $\Rightarrow dxy(x+y)\vdots x^2+xy+y^2,$ note that $\gcd(x,y)=1$ so we can easily prove that$$\gcd(xy,x^2+xy+y^2)=\gcd(x+y,x^2+xy+y^2)=1$$Hence $d\vdots x^2+xy+y^2 \Rightarrow d>3xy$ (since $x>y$) $\Rightarrow (k-m)^3=d^3(x-y)^3=d^2d(x-y)^3>d^23xy=3km.$