2022 AMC 12A Problem 14

Let $u=\log 2.$ Then $\log 5=1-u,$ $\log 20=1+u,$ $\log 8=3u,$ and $\log 0.25=-2u.$ With $a=1-u,\ b=1+u,$ we have $a+b=2$ and $ab=1-u^2,$ so $$\begin{gathered} a^3+b^3 \\ =(a+b)\big((a+b)^2-3ab\big) \\ =2\big(4-3(1-u^2)\big) \\ =2+6u^2. \end{gathered}$$ The last term is $(3u)(-2u)=-6u^2,$ so the total is $2+6u^2-6u^2=2.$ Thus, the correct answer is C.