Find all real numbers a that satisfy \frac{1}{64a^3 + 7} - 7 = 0.
we are trying to find the real numbers that satisfy 1/(64a^3)-7=0
first we add 7 to both sides 1/(64a^3)=7
then we multiply to get 1=488a^3+49
subtract 49 to get -48=488a^3
-48/488=a^3
-3/28=a^3
so we have cuberoot(-3/28)=a
hope this helps
