[Algebra]
Problem: Determine all points $(x, y)$ where the two curves $y=\log _{10}\left(x^{4}\right)$ and $y=\left(\log _{10} x\right)^{3}$ intersect.

Solution: To determine the points of intersection, we equate $y$ values of the two curves and obtain $\log _{10}\left(x^{4}\right)=\left(\log _{10} x\right)^{3}$.

Since $\log _{10}\left(a^{b}\right)=b \log _{10} a$, the equation becomes $4 \log _{10} x=\left(\log _{10} x\right)^{3}$.

We set $u=\log _{10} x$ and so the equation becomes $4 u=u^{3}$, or $u^{3}-4 u=0$.

We can factor the left side as $u^{3}-4 u=u\left(u^{2}-4\right)=u(u+2)(u-2)$.

Therefore, $u(u+2)(u-2)=0$, and so $u=0$ or $u=-2$ or $u=2$.

Therefore, $\log _{10} x=0$ or $\log _{10} x=-2$ or $\log _{10} x=2$.

Therefore, $x=1$ or $x=\frac{1}{100}$ or $x=100$.

Finally, we must calculate the $y$-coordinates of the points of intersection. Since one of the original curves is $y=\left(\log _{10} x\right)^{3}$, we can calculate the corresponding values of $y$ by using the fact that $y=u^{3}$.

The corresponding values of $y$ are $y=0^{3}=0$ and $y=(-2)^{3}=-8$ and $y=2^{3}=8$.

Therefore, the points of intersection are $(1,0),\left(\frac{1}{100},-8\right)$ and $(100,8)$.

Final answer: $(1,0),(\frac{1}{100},-8),(100,8)$