Question

If x=2+under root3 Then Show that x³ + 1/x³=52​

Answer 1

Step-by-step explanation:

Given :

To prove $x^3 + \frac{1}{x^3} = 52$, if $x = 2 + \sqrt{3}$

Solution :

We know that,

$x = 2 + \sqrt{3}$,.

Then,

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

By taking conjugate,

We get,

$\frac{1}{ 2 + \sqrt{3}}\times\frac{2 - \sqrt{3}}{2 - \sqrt{3}}$

⇒ $\frac{2 + \sqrt{3}}{(2 + \sqrt{3})(2 - \sqrt{3})}$

By using the identity (a + b)(a - b) = a² - b²,

By substituting a = 2 , b = √3,

We get,

$\frac{2 - \sqrt{3}}{(2)^2-(\sqrt{3})^2}}=\frac{2- \sqrt{3}}{4- 3}=2-\sqrt{3}$

∴ $\frac{1}{x} = 2 - \sqrt{3}$

Hence,

By using $(a + b)^3 = a^3 + b^3 + 3a^2b + 3ab^2$

& $(a - b)^3 = a^3 - 3a^2b + 3ab^2 - b^3$

Hence,

$(a + b)^3 + (a - b)^3 = 2a^3 + 6ab^2$

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

Here, a = 2 , b = √3

⇒ $(2 + \sqrt{3} )^3 + (2 - \sqrt{3})^3 = 2(2^3) + 6(2)((\sqrt{3})^2) = 2(8) + 6(2)(3) = 16 + 36 = 52$

Hence, proved.

Answer 2

I hope it will help you