Solve c^6-20c^3+64=0

$${ c }^{ 6 } -20 { c }^{ 3 } +64=0$$

Solve for c

$$c=\sqrt[3]{4}\approx 1.587401052$$

$$c=2^{\frac{4}{3}}\approx 2.5198421$$

Steps Using the Quadratic Formula

Substitute t for c^{3}.

$$t^{2}-20t+64=0$$

All equations of the form ax^{2}+bx+c=0 can be solved using the quadratic formula: \frac{-b±\sqrt{b^{2}-4ac}}{2a}. Substitute 1 for a, -20 for b, and 64 for c in the quadratic formula.

$$t=\frac{-\left(-20\right)±\sqrt{\left(-20\right)^{2}-4\times 1\times 64}}{2}$$

Do the calculations.

$$t=\frac{20±12}{2}$$

Solve the equation t=\frac{20±12}{2} when ± is plus and when ± is minus.

$$t=16 t=4$$

Since c=t^{3}, the solutions are obtained by evaluating c=\sqrt[3]{t} for each t.

$$c=2\sqrt[3]{2} c=\sqrt[3]{4}$$