Question

$x = \sqrt[3]{2 + \sqrt{3} }$
then the value of
x^3+1/x^3 =​

Answer 1

Answer:

Step-by-step explanation:

We have,

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

$\implies\,{x}^{3}=2+\sqrt{3}\,\,\,\,\,\,\,\,\,\,\,...(1)$

$\implies\,\dfrac{1}{{x}^{3}}=\dfrac{1}{2+\sqrt{3}}$

$\implies\,\dfrac{1}{{x}^{3}}=\dfrac{\left(2-\sqrt{3}\right)}{\left(2+\sqrt{3}\right)\left(2-\sqrt{3}\right)}$

$\implies\,\dfrac{1}{{x}^{3}}=\dfrac{2-\sqrt{3}}{\left(2\right)^2-\left(\sqrt{3}\right)^2}$

$\implies\,\dfrac{1}{{x}^{3}}=\dfrac{2-\sqrt{3}}{4-3}$

$\implies\,\dfrac{1}{{x}^{3}}=\dfrac{2-\sqrt{3}}{1}$

$\implies\,\dfrac{1}{{x}^{3}}=2-\sqrt{3}\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(2)$

Adding (1) and (2),

$\implies\,{x}^{3}+\dfrac{1}{{x}^{3}}=2+\sqrt{3}+2-\sqrt{3}$

$\implies\,{x}^{3}+\dfrac{1}{{x}^{3}}=2+2$

$\implies\,{x}^{3}+\dfrac{1}{{x}^{3}}=4$