Question

show that
${x}^{3 }-6x - 6 = 0$
if
$x = \sqrt[3]{2} + \sqrt[3]{4}$

Answer 1

$x = \sqrt[3]{2} + \sqrt[3]{4} \\ x - \sqrt[3]{2} = \sqrt[ 3 ]{4} \\ cubing \: both \: sides \\ {(x - \sqrt[3]{2}) }^{3} = 4 \\ {x}^{3} - 2 - 3 \times x \times \sqrt[3]{2} \times \sqrt[3]{4} = 4 \\ {x}^{3} - 6 - 6x = 0$
hence proved

Answer 2

x³−6x−6=0
if
$x = \sqrt[3]{2} + \sqrt[3]{4} ​$

$\begin{lgathered}x = \sqrt[3]{2} + \sqrt[3]{4} \\ x - \sqrt[3]{2} = \sqrt[ 3 ]{4} \\ cubing \: both \: sides \\ {(x - \sqrt[3]{2}) }^{3} = 4 \\ {x}^{3} - 2 - 3 \times x \times \sqrt[3]{2} \times \sqrt[3]{4} = 4 \\ {x}^{3} - 6 - 6x = 0\end{lgathered}$

HENCE PROVED

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