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.
+15000 1.
\( \dfrac{1}{64a^3 + 7} - 7 = 0\\ 7\cdot (64a^3+7)=1\\ 7\cdot 64a^3+48=0\\ a=(-\dfrac{48}{7\cdot 64})^{\frac{1}{3}}\\ a=(-\dfrac{3}{7\cdot 4})^{\frac{1}{3}}\\ \color{blue}a=-0.474957 \)
2.
\( \dfrac{a}{3} + 1 = \dfrac{a + 3}{a} + 1\\ \frac{a}{3}=\dfrac{a+3}{a}\\ a^2-3a-9=0\\ a=\dfrac{3}{2}\pm (\dfrac{9}{4}+9)^{\frac{1}{2}}\\ \color{blue} a\in \{-1.854,\ 4.854\}\)
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