Equation

**bhitroofen01** · April 3rd 2010, 04:05 AM

Hi people,

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

Can you help me please???

**earboth** · April 3rd 2010, 04:16 AM

Originally Posted by bhitroofen01

Hi people,

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

Can you help me please???

Transfer all terms to the RHS except x:

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

$(\sqrt[3]{x}-1)^3=54$ Now calculate the 3rd root of both sides of the equation:

$\sqrt[3]{x}-1=\sqrt[3]{54}=3\sqrt[3]{2}$

$\sqrt[3]{x}=1+3\sqrt[3]{2}$ Now get rid of the cube root:

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

**Prove It** · April 3rd 2010, 04:18 AM

Originally Posted by bhitroofen01

Hi people,

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

Can you help me please???

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

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

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

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

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

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

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

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

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