• Define $gcd$ of $M,N$
• Use the theorem $lcm(a,b) \times \gcd(a,b) = ab$
• Use divisibility and co-prime numbers in your advantage.

Let the gcd of $M$ and $N$ be $d$.

Let $M=dm$ and $N=dn$.
As $d$ is gcd, $gcd(m,n)=1$

So as given $lcm(M,N)=M^2-N^2+MN$

$lcm(M,N)=d^2m^2-d^2n^2+d^2mn=d^2(m^2-n^2+mn)$

We know, $gcd(a,b) \times lcm(a,b)=ab$

Using this,

$lcm(M,N) \times gcd(M,N)=MN$

$d^2(m^2-n^2+mn) \times d = dm \times dn$

$d^3(m^2-n^2+mn)=d^2mn$

$d(m^2-n^2+mn)=mn$

$dm^2-dn^2+dmn=mn$

$mn(d-1)=d(n^2-m^2)=d(n-m)(n+m)$

Note that $d$ is a divisor of Left hand side

As $gcd(d,d-1)=1$ and $d | mn(d-1)$ we have $d | mn$

As $gcd(m,n)=1$ 

So, $gcd(m,n-m)=1$

Also $gcd(m,n+m)=1$

So, $gcd(m,(n-m)(n+m))=1$

Similarly $gcd(n,(n-m)(n+m))=1$

So, $gcd(mn,(n-m)(n+m))=1$

which implies $mn | d$

We have $d | mn$ and $mn | d$ so $d=mn$

Now, $MN=dm \times dn =d^2mn=d^3$ which is a cubic number.