Solve (-5+8a^3)^2=I

$$( - 5 + 8 a ^ { 3 } ) ^ { 2 } = I$$

Solve for I

$$I=\left(8a^{3}-5\right)^{2}$$

Solution Steps

Use binomial theorem \left(p+q\right)^{2}=p^{2}+2pq+q^{2} to expand \left(-5+8a^{3}\right)^{2}.

$$25-80a^{3}+64\left(a^{3}\right)^{2}=I$$

To raise a power to another power, multiply the exponents. Multiply 3 and 2 to get 6.

$$25-80a^{3}+64a^{6}=I$$

Swap sides so that all variable terms are on the left hand side.

$$I=25-80a^{3}+64a^{6}$$

Solve for a (complex solution)

$$a\in \frac{\sqrt[3]{\sqrt{I}+5}}{2},\frac{e^{\frac{4\pi i}{3}}\sqrt[3]{\sqrt{I}+5}}{2},\frac{e^{\frac{2\pi i}{3}}\sqrt[3]{\sqrt{I}+5}}{2},\frac{e^{\frac{4\pi i}{3}}\sqrt[3]{-\sqrt{I}+5}}{2},\frac{\sqrt[3]{-\sqrt{I}+5}}{2},\frac{e^{\frac{2\pi i}{3}}\sqrt[3]{-\sqrt{I}+5}}{2}$$

Solve for a

$$a=\frac{\sqrt[3]{-\sqrt{I}+5}}{2}$$

$$a=\frac{\sqrt[3]{\sqrt{I}+5}}{2}\text{, }I\geq 0$$