•  Expand. Use $(m+n+p)^3$
•  From $(m+n+p)(\frac{1}{m}+\frac{1}{n}+\frac{1}{p})=1$, we can show $(m+n+p)^3-m^3-n^3-p^3=0$
•  Factorize and assume WLOG $n+p=0$

$(m+n+p)(\frac{1}{m}+\frac{1}{n}+\frac{1}{p})=1$

$\implies \frac{1}{m}+\frac{1}{n}+\frac{1}{p}=\frac{1}{m+n+p}$

$\implies \frac{mn+mp+np}{mnp}=\frac{1}{m+n+p}$

$\implies (m+n+p)(mn+mp+nm)=mnp$

$\implies m^2n+mn^2+m^2p+mp^2+n^2p+np^2+3mnp=mnp$

$\implies m^2n+mn^2+m^2p+mp^2+n^2p+np^2+2mnp=0$

$\implies 3(m^2n+mn^2+m^2p+mp^2+n^2p+np^2)+6mnp=0$

$\implies (m+n+p)^3-m^3-n^3-p^3=0$

$\implies ((m+n+p)^3-m^3)-(n^3+p^3)=0$ 

$\implies (n+p)((n+p)^2 + m(n+p) +m^2) -(n+p)(n^2-np+p^2)=0$

$\implies (n+p)(m^2+n^2+p^2+2(mn+np+mp)+m^2+mn+mp+m^2-n^2+np-p^2)=0$

$\implies (n+p)(3m^2+3mn+3pn+3pm)=0$

$\implies (n+p)(m+p)(m+n)=0$

We can assume without loss of generality that

\[n+p=0\implies n=-p\]

Now, for the given term

\[\frac{1}{(m+(-p)+p)^{2023}}-\frac{1}{m^{2023}}-\frac{1}{(-p)^{2023}}-\frac{1}{p^{2023}}=\frac{1}{m^{2023}}-\frac{1}{m^{2023}}+\frac{1}{p^{2023}}-\frac{1}{p^{2023}}=0\]

So, the answer is 0.