\text{If }m+n+o=\mathrm{15 }\text{and }mn+no+om=75\text{, then determine the value of}{\mathrm{ m}}^3+n^3+o^3-3mno.

The correct option is 0.

To determine the value of the algebraic expression $m^3+n^3+o^3-3mno$, we can use a well-known algebraic identity:
$m^3+n^3+o^3-3mno=(m+n+o)(m^2+n^2+o^2-mn-n-om)$

We are given the following values:
1) $m+n+o=15$
2) $mn+n+om=75$

We first need to find the value of $m^2+n^2+o^2$. We can find this by squaring the equation $m+n+o=15$:
${(m+n+o)}^2={15}^2$

Expanding the left side of the equation, we get:
$m^2+n^2+o^2+2(mn+n+om)=225$

Substitute the value of $mn+n+om=75$ into this expression:
$m^2+n^2+o^2+2\left(75\right)=225$
$m^2+n^2+o^2+150=225$
$m^2+n^2+o^2=225-150$
$m^2+n^2+o^2=75$

Now, we can substitute all the known values back into our original identity:
$m^3+n^3+o^3-3mno=(m+n+o)\left[\right(m^2+n^2+o^2)-(mn+n+om\left)\right]$
$m^3+n^3+o^3-3mno=15·(75-75)$
$m^3+n^3+o^3-3mno=15·0$
$m^3+n^3+o^3-3mno=0$

Thus, the final evaluated value is indeed 0.