$(m+n+p)(mn+np+pm)=mnp \to (m+n)(n+p)(p+m)=0$ Let $m+n=0$ then $\frac 1{(m+n+p)^{2023}} -\frac 1{m^{2023}} -\frac 1{n^{2023}} -\frac 1{p^{2023}}=0$

After u simplify get $(m+n)(mn+p^2+pm+pn)=0$ Solve for two cases thus we get possible values to be $0$

From the equiation we have that $$(m+n+p)(mn+np+pm)-mnp=0$$Therefore, $$(m+n)(n+p)(p+m)=0.$$The last equality implies that at least one of the three factors is equal to zero. Suppose without loss of generality that $m+p=0$. Then, $m^{2023}=-p^{2023}$, so we conclude that $$\frac{1}{m^{2023}}+\frac{1}{p^{2023}}=0.$$Therefore, $$\frac 1{(m+n+p)^{2023}} -\frac 1{m^{2023}} -\frac 1{n^{2023}} -\frac 1{p^{2023}} =\frac{1}{n^{2023}}-\frac{1}{n^{2023}}=0.$$We conclude that the only value that the requested expression can take is zero.

Claim: $\left(m+n+p\right)\left(\frac 1m + \frac 1n + \frac1p\right) =1 \implies (m+n)(n+p)(p+m)=0.$ Proof: $\left(m+n+p\right)\left(\frac 1m + \frac 1n + \frac1p\right)=3+\frac{m}{n}+\frac{m}{p}+\frac{n}{m}+\frac{n}{p}+\frac{p}{m}+\frac{p}{n}=1$, note that this implies, $(m+n+p)(mn+mp+np)-mnp=(m+n)(m+p)(n+p)=0.$ Now, WLOG, assume $m+p=0 \implies m=-p$, hence, $\frac 1{(m+n+p)^{2023}} -\frac 1{m^{2023}} -\frac 1{n^{2023}} -\frac 1{p^{2023}}=\frac{1}{(m+n+p)^{2023}}-\frac{1}{p^{2023}}=\frac{1}{p^{2023}}-\frac{1}{p^{2023}}=\boxed{0}.$

(m + n + p)(1/m + 1/n + 1/p) = 1 After a bit of simplification, we get (m + n + p)(mn + np + pm) - mnp = 0 m²n + mn² + n²p + np² + pm² + p²m + 2mnp = 0 m²(n + p) + np(n + p) + mn(n + p) + pm(n + p) = 0 (n + p)(m² + mn + np + pm) = 0 (m + n)(n + p)(p + m) = 0 Now, by Without Loss of Generality we assume that (m + n) = 0 i.e. m = -n Therefore, 1/(m + n + p)^2023 - 1/m^2023 - 1/n^2023 - 1/p^2023 = 1/p^2023 - 1/p^2023 = 0