What's the remainder when $x^{7} + x^{27} + x^{47} +x^{67} + x^{87}$ is divided by $x ^ 3 - x$

Modulo $x^3-x$ we have $x^3=x$ so $x^{2n+1}=x$ for any integer $n\ge 0$ (a trivial induction exercise). Hence your sum leaves remainder $5x$.

Let your $87$-th degree polynomial be $P(x)$. You're to find $Q(x)=ax^2+bx+c$ where $$ P(x)=(x^3-x)D(x)+Q(x). $$ Then: $$ 5=P(1)=0\cdot D(1)+Q(1)\implies Q(1)=5;\\ 0=P(0)=0\cdot D(0)+Q(0)\implies Q(0)=0;\\ -5=P(-1)=0\cdot D(-1)+Q(-1)\implies Q(-1)=-5. $$ From $Q(0)=0$, it is easy to see $c=0$. Using this and the other 2 conditions, we have: $$ a+b=5,\quad a-b=-5\implies a=0,\quad b=5. $$ In sum $Q(x)=5x$.

$xf(x^2)\,\bmod\, x(x^2\!-\!1)\, =\, x\,(\overbrace{f(\color{#c00}{x^2})\,\bmod\, x^2\!-\!1}^{\color{#c00}{\Large x^2\ \equiv\,\ 1}})\, =\, xf(\color{#c00}{ 1})\, =\, 5x$