Calculate $x^3 + \frac{1}{x^3}$ Question
$x^2 + \frac{1}{x^2}=34$ and $x$ is a natural number. Find the value of $x^3 +  \frac{1}{x^3}$ and choose the correct answer from the following options:

*

*198

*216

*200

*186


What I have did yet
I tried to find the value of $x + \frac{1}{x}$. Here are my steps to do so:
$$x^2 + \frac{1}{x^2} = 34$$
$$\text{Since}, (x+\frac{1}{x})^2 = x^2 + 2 + \frac{1}{x^2}$$
$$\Rightarrow (x+\frac{1}{x})^2-2=34$$
$$\Rightarrow (x+\frac{1}{x})^2=34+2 = 36$$
$$\Rightarrow x+\frac{1}{x}=\sqrt{36}=6$$
I have calculated the value of $x+\frac{1}{x}$ is $6$. I do not know what to do next. Any help will be appreciated. Thank you in advanced!

You wrote
$$(x+\frac{1}{x}) = x^2 + 2 + \frac{1}{x^2}.$$
But it should read
$$(x+\frac{1}{x})^2 = x^2 + 2 + \frac{1}{x^2}.$$
Then we obtain $x+\frac{1}{x}=6.$
From $x^2 + \frac{1}{x^2}=34$ we obtain
$(1) \quad x^3+\frac{1}{x}=34 x$
and
$(2) \quad x+\frac{1}{x^3}=\frac{34}{x}.$
If we add (1) and (2) we get
$$x^3+\frac{1}{x^3}=33(x+\frac{1}{x}).$$
Ans so
$$x^3+\frac{1}{x^3}=6 \cdot 33=198.$$
Remark : if $x^2 + \frac{1}{x^2}=34$, then $x$ can not be a natural number !
