Hi people,
I don't know how to calculate $\displaystyle (\sqrt[3]{x}-1)^3-54=0$???
Can you help me please???
Transfer all terms to the RHS except x:
$\displaystyle (\sqrt[3]{x}-1)^3-54=0$
$\displaystyle (\sqrt[3]{x}-1)^3=54$ Now calculate the 3rd root of both sides of the equation:
$\displaystyle \sqrt[3]{x}-1=\sqrt[3]{54}=3\sqrt[3]{2}$
$\displaystyle \sqrt[3]{x}=1+3\sqrt[3]{2}$ Now get rid of the cube root:
$\displaystyle x=(1+3\sqrt[3]{2})^3$
$\displaystyle (\sqrt[3]{x} - 1)^3 - 54 = 0$
$\displaystyle (\sqrt[3]{x} - 1)^3 = 54$
$\displaystyle \sqrt[3]{x} - 1 = \sqrt[3]{54}$
$\displaystyle \sqrt[3]{x} - 1 = \sqrt[3]{27\cdot 2}$
$\displaystyle \sqrt[3]{x} - 1 = \sqrt[3]{27} \cdot \sqrt[3]{2}$
$\displaystyle \sqrt[3]{x} - 1 = 3\sqrt[3]{2}$
$\displaystyle \sqrt[3]{x} = 3\sqrt[3]{2} + 1$
$\displaystyle x = (3\sqrt[3]{2} + 1)^3$.