If $x = 2^{\frac{1}{3}} + 2^{\frac{2}{3}}$, show that $x^3 - 6x = 6$.

Given:

$x = 2^{\frac{1}{3}} + 2^{\frac{2}{3}}$

To do: 

We have to show that $x^3 - 6x = 6$.

Solution:

We know that,

$(a^{m})^{n}=a^{m n}$

$a^{m} \times a^{n}=a^{m+n}$

$a^{m} \div a^{n}=a^{m-n}$

$a^{0}=1$

Therefore,

$x=2^{\frac{1}{3}}+2^{\frac{2}{3}}$

Cubing both sides, we get,

$x^{3}=(2^{\frac{1}{3}}+2^{\frac{2}{3}})^{3}$

$=(2^{\frac{1}{3}})^3+(2^{\frac{2}{3}})^{3}+3 \times 2^{\frac{1}{3}} \times 2^{\frac{2}{3}}(2^{\frac{1}{3}}+2^{\frac{2}{3}})$

$=2+4+3 \times 2^{\frac{1}{3}+\frac{2}{3}} \times x$

$=6+3 \times 2 \times x$

$=6 x+6$

$\Rightarrow x^{3}-6 x=6$

Hence proved.

Tutorialspoint

Updated on: 10-Oct-2022

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