Find all real numbers a that satisfy \frac{1}{64a^3 + 7} - 7 = 0.
\(\frac{1}{64a^3 + 7} - 7 = 0 \)
1/[64a^3 + 7] = 7
1 = 7 [ 64a^3 + 7]
1 = 448a^3 + 49
-48 = 448 a^3
[ -48 /448 ] = a^3
-3/28 = a^3
a = - ∛ ( 3/28)
