Solve for x if $\sqrt[3]{x^2} + \sqrt[3]{x} - 20 = 0$.

Use the CALC function

X = 64 will result to zero

Answer: x = 64
 

Another Solution
$\sqrt[3]{x^2} + \sqrt[3]{x} - 20 = 0$

$(\sqrt[3]{x})^2 + \sqrt[3]{x} - 20 = 0$
 

Let $z = \sqrt[3]{x}$
$z^2 + z - 20 = 0$

$z = 4 ~ \text{ and } ~ -5$
 

For z = 4
$4 = \sqrt[3]{x}$

$x = 64$

Check: $\sqrt[3]{64^2} + \sqrt[3]{64} - 20 = 0$   ←   okay!
 

For z = -5
$-5 = \sqrt[3]{x}$

$x = -125$

Check: $\sqrt[3]{(-125)^2} + \sqrt[3]{-125} - 20 = 0$   ←   okay!
 

Answer: x = 64 and x = -125