# Equation

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• Apr 3rd 2010, 04:05 AM
bhitroofen01
Equation
Hi people,

I don't know how to calculate $\displaystyle (\sqrt[3]{x}-1)^3-54=0$???

Can you help me please???
• Apr 3rd 2010, 04:16 AM
earboth
Quote:

Originally Posted by bhitroofen01
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:

$\displaystyle x=(1+3\sqrt[3]{2})^3$
• Apr 3rd 2010, 04:18 AM
Prove It
Quote:

Originally Posted by bhitroofen01
Hi people,

I don't know how to calculate $\displaystyle (\sqrt[3]{x}-1)^3-54=0$???

Can you help me please???

$\displaystyle (\sqrt[3]{x} - 1)^3 - 54 = 0$

$\displaystyle (\sqrt[3]{x} - 1)^3 = 54$

$\displaystyle \sqrt[3]{x} - 1 = \sqrt[3]{54}$

$\displaystyle \sqrt[3]{x} - 1 = \sqrt[3]{27\cdot 2}$

$\displaystyle \sqrt[3]{x} - 1 = \sqrt[3]{27} \cdot \sqrt[3]{2}$

$\displaystyle \sqrt[3]{x} - 1 = 3\sqrt[3]{2}$

$\displaystyle \sqrt[3]{x} = 3\sqrt[3]{2} + 1$

$\displaystyle x = (3\sqrt[3]{2} + 1)^3$.