Question

x= ³√(2+√3 ,then x³+1/x³ is​

Answer 1

Answer :-

$\implies\sf x = \sqrt[3]{2+\sqrt3}$

$\implies\sf x = (2+\sqrt3)^\frac{1}{3}$

Calculating value of x³ :-

$\implies\sf x^3 = \left[(2+\sqrt3)^\frac{1}{3}\right]^3$

$\implies\sf x^3 = (2+\sqrt3)^{\not3\times \frac{1}{\not3}}$

$\implies\sf x^3 = 2 + \sqrt3$

Calculating value of 1/x³ :-

$\implies\sf \dfrac{1}{x^3} = \dfrac{1}{2+\sqrt3}$

$\implies\sf \dfrac{1}{x^3} = \dfrac{1}{2+\sqrt3} \times \dfrac{2-\sqrt3}{2-\sqrt3}$

$\implies\sf \dfrac{1}{x^3} = \dfrac{2-\sqrt3}{(2+\sqrt3)(2-\sqrt3)}$

$\implies\sf \dfrac{1}{x^3} = \dfrac{2-\sqrt3}{2^2 - (\sqrt3)^2}$

$\implies\sf \dfrac{1}{x^3} = \dfrac{2-\sqrt3}{4-3}$

$\implies\sf \dfrac{1}{x^3} = \dfrac{2-\sqrt3}{1}$

$\implies\sf \dfrac{1}{x^3} = 2-\sqrt3$

Calculating value of x³ + 1/x³ :-

$\implies\sf x^3 + \dfrac{1}{x^3} = 2+\sqrt3+2-\sqrt3$

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

Additional information :-

$\bf Algebraic\:Identities\:\bigstar\\\\1)\bf\:(A+B)^{2} = A^{2} + 2AB + B^{2}\\\\2)\sf\: (A-B)^{2} = A^{2} - 2AB + B^{2}\\\\3)\bf\: A^{2} - B^{2} = (A+B)(A-B)\\\\4)\sf\: (A+B)^{2} = (A-B)^{2} + 4AB\\\\5)\bf\: (A-B)^{2} = (A+B)^{2} - 4AB\\\\6)\sf\: (A+B)^{3} = A^{3} + 3AB(A+B) + B^{3}\\\\7)\bf\:(A-B)^{3} = A^{3} - 3AB(A-B) + B^{3}\\\\8)\sf\: A^{3} + B^{3} = (A+B)(A^{2} - AB + B^{2})$

Answer 2

Given

• $\sf x=\sqrt[3]{2+\sqrt{3}}$

To Find

• $\sf x³+\dfrac{1}{x³}$

Solution

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

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

On Cubing both sides

$~~~~~~:~~\implies\sf {x}^{3} = \{( 2+\sqrt{3} )^{ \frac{1}{3} } \}^{3}$

$~~~~~~:~~\implies\sf {x}^{3} = \{( 2+\sqrt{3} )^{ 3 \times \frac{1}{3} } \}$

$~~~~~~:~~ \pink{\implies\sf {x}^{3} = ( 2+\sqrt{3} ) }$

$~~~~~~:~~\implies\sf \dfrac{1}{x³} =\dfrac{1}{ ( 2+\sqrt{3} )}$

${~~~~~~:~~\implies\sf \dfrac{1}{x³} =\dfrac{1}{ 2+\sqrt{3} } \times \dfrac{2 - \sqrt{3} }{2 - \sqrt{3} } }$

${~~~~~~:~~\implies\sf \dfrac{1}{x³} =\dfrac{2 - \sqrt{3} }{ {2}^{2} - (\sqrt{3} )^{2} } }$

${~~~~~~:~~\implies\sf \dfrac{1}{x³} =\dfrac{2 - \sqrt{3} }{ 4 - 3 } }$

${~~~~~~:~~\implies\sf \green{\dfrac{1}{x³} =2 - \sqrt{3} } }$

Now

${~~~~~~:~~\implies\sf x³+\dfrac{1}{x³}= (2+\sqrt{3})+(2-\sqrt{3})}$

${~~~~~~:~~\implies\sf x³+\dfrac{1}{x³}= 2 + 2\: \: \: \xcancel{+\sqrt{3}} \: \: \: \: \xcancel{-\sqrt{3}}}$

${~~~~~~~~~~~~~~~~~~~~\boxed{\bf x³+\dfrac{1}{x³}=4}}$

$\begin{gathered}\\\\\begin{gathered}\begin{gathered}\begin{gathered}\begin{gathered} \bigstar \: \underline{\bf{\blue{More \: Useful \: Formula}}}\\ {\boxed{\begin{array}{cc}\dashrightarrow \sf(a + b)^{2} = {a}^{2} + {b}^{2} + 2ab \\\dashrightarrow \sf(a - b)^{2} = {a}^{2} + {b}^{2} - 2ab \\\dashrightarrow \sf(a + b)(a - b) = {a}^{2} - {b}^{2} \\\dashrightarrow \sf(a + b) ^{3} = {a}^{3} + b^{3} + 3ab(a + b) \\ \dashrightarrow\sf(a - b) ^{3} = {a}^{3} - b^{3} - 3ab(a - b) \\ \dashrightarrow\sf a ^{3} + {b}^{3} = (a + b)(a ^{2} + {b}^{2} - ab) \\\dashrightarrow \sf a ^{3} - {b}^{3} = (a - b)(a ^{2} + {b}^{2} + ab \\\dashrightarrow \sf{a²+b²=(a+b)²-2ab}\\ \end{array}}}\end{gathered}\end{gathered}\end{gathered}\end{gathered}\end{gathered}$