Question

if x=2+√3 find tge calue of x^3+1÷x^3​

Answer 1

Step-by-step explanation:

here is your answer pls follow me if it is correct

Answer 2

Answer:

52

Step-by-step explanation:

Given :

$\large x=2+\sqrt3$

We have to find value of   $\large x^3+\dfrac{1}{x^3}$

First take the reciprocal of x

$\large \dfrac{1}{x}= \dfrac{1}{2+\sqrt3}$

Now rationalize it.

$\large \dfrac{1}{x}= \dfrac{1}{2+\sqrt3}\times \dfrac{2-\sqrt3}{2-\sqrt3} \\\\\\\large \dfrac{1}{x}=\dfrac{2-\sqrt3}{4-3} \\\\\\ \large \dfrac{1}{x}= 2-\sqrt3$

Now add both

$\large x+\dfrac{1}{x}=2+\sqrt3+2-\sqrt3\\\\\\\large x+\dfrac{1}{x}=4$

Now cubic on both side we get

$\large (x+\dfrac{1}{x})^3=(4)^3\\\\\\\large \text{Using indentity (a+b)^3=a^3+b^3+3ab(a+b) }\\\\\\\large (x)^3+(\dfrac{1}{x})^3+3(4-3)(4)=64\\\\\\\large (x)^3+(\dfrac{1}{x})^3+12=64\\\\\\\large (x)^3+(\dfrac{1}{x})^3=64-12\\\\\\\large (x)^3+(\dfrac{1}{x})^3=52$

Thus we get answer 52