1. Find all real numbers a that satisfy \frac{1}{64a^3 + 7} - 7 = 0.
2. Find all values of a that satsify the equation \frac{a}{3} + 1 = \frac{a + 3}{a} + 1.
1. A = -2
Expand out terms of the left hand side
1-7a³-49=-a³
-7a³-48=-a³
Taking out minus common from both sides and cancel them
-(7a³+48)=-(a³)
7a³+48=a³
Add -a³-48 on both sides
7a³+48-a³-48=a³-a³-48
6a³=-48
Divide both sides with 6
(6÷6)a³=(-48÷6)
a³=-8
Write -8 in the expansion form which is (-2)×(-2)×(-2)=-8
a³=(-2)³
From above it says that
a=-2
2. Sorry for this one I do not know. But I hope the answer up top will do something to satisfy you.