That's not enough!

$\begin{aligned} \left(x+\frac{1}{x} \right)^3 & = x^3 +3x + \frac{3}{x} + \frac{1}{x^3} = 18+ 3 \left(x+\frac{1}{x} \right) \\ \left(x+\frac{1}{x} \right)^3 & - 3 \left(x+\frac{1}{x} \right) - 18 = 0 \\ \Rightarrow x+\frac{1}{x} & = 3 \end{aligned}$

$\begin{aligned} \left(x+\frac{1}{x} \right)^2 & = x^2 + 2 + \frac{1}{x^2} = 9 \\ \Rightarrow x^2 + \frac{1}{x^2} & = 7 \\ \Rightarrow x^4 + \frac{1}{x^4} & = 7^2 - 2 = 47 \\ \Rightarrow x^8 + \frac{1}{x^8} & = 47^2 - 2 = 2207 \end{aligned}$

$\begin{aligned} \left(x^3+\frac{1}{x^3} \right)\left(x^2+\frac{1}{x^2} \right) & = x^{5} + x + \frac{1}{x} + \frac{1}{x^{5}} \\ 18(7) & = x^{5} + \frac{1}{x^{5}} + 3 \\ \Rightarrow x^{5} + \frac{1}{x^{5}} & = 123 \\ \left(x^{8}+\frac{1}{x^{8}} \right)\left(x^3+\frac{1}{x^3} \right) & = x^{11} + x^5 + \frac{1}{x^5} + \frac{1}{x^{11}} \\ 2207(18) & = x^{11} + \frac{1}{x^{11}} + 123 \\ \Rightarrow x^{11} + \frac{1}{x^{11}} & = 39726 - 123 = \boxed{39603} \end{aligned}$

First note that $\left(x + \dfrac{1}{x}\right)^{2} = x^{2} + \dfrac{1}{x^{2}} + 2,$ and that

$\left(x + \dfrac{1}{x}\right)\left(x^{2} + \dfrac{1}{x^{2}}\right) = x^{3} + \dfrac{1}{x^{3}} + \left(x + \dfrac{1}{x}\right).$

Letting $a = x + \dfrac{1}{x},$ this last equation then becomes

$a(a^{2} - 2) = 18 + a \Longrightarrow a^{3} - 3a - 18 = 0 \Longrightarrow (a - 3)(a^{2} + 3a + 6) = 0.$

Since the discriminant of $a^{2} + 3a + 6$ is $-15 \lt 0,$ and since with $x$ real we must have $a$ real, we see that $a = x + \dfrac{1}{x} = 3.$ Thus

$x^{2} + \dfrac{1}{x^{2}} = 7 \Longrightarrow x^{4} + \dfrac{1}{x^{4}} = \left(x^{2} + \dfrac{1}{x^{2}}\right)^{2} - 2 = 49 - 2 = 47.$

Now $\left(x^{3} + \dfrac{1}{x^{3}}\right)\left(x^{4} + \dfrac{1}{x^{4}}\right) = x^{7} + \dfrac{1}{x^{7}} + \left(x + \dfrac{1}{x}\right)$

$\Longrightarrow x^{7} + \dfrac{1}{x^{7}} = 18*47 - 3 = 843.$

Next, $\left(x^{7} + \dfrac{1}{x^{7}}\right)\left(x^{4} + \dfrac{1}{x^{4}}\right) = x^{11} + \dfrac{1}{x^{11}} + \left(x^{3} + \dfrac{1}{x^{3}}\right)$

$\Longrightarrow x^{11} + \dfrac{1}{x^{11}} = 843*47 - 18 = \boxed{39603}.$

Mess-free calculator trick.

If $x$ is a real number, then it must be positive, since negative values of $x$ cannot result in a positive $x^3 + 1/{x^3}$ ; and $x = 0$ would probably spawn a black hole.

Since $x$ is positive, we can safely substitute $x = \exp(a)$ . By definition of $\cosh$ , note that

$x^3 + \frac1{x^3} = 2\cosh 3a = 18$

$a = \frac13 \cosh^{-1} 9$

$x^{11} + \frac1{x^{11}} = 2\cosh 11a = 2\cosh\left( \frac{11}3 \cosh^{-1} 9\right)$

Punch some calculator buttons to get 39603

It is given that $x^3+\dfrac{1}{x^3} = 18.$

Let us assume that $x+\dfrac{1}{x} = a.$

Now, $(x+\dfrac{1}{x})^3 = x^3+3(x+\dfrac{1}{x})+\dfrac{1}{x^3}.$

Putting $a=x+\dfrac{1}{x}$ and $x^3+\dfrac{1}{x^3}=18$ we get ,

$a^3=18+3a \Rightarrow a^3-3a-18=0 \Rightarrow (a-3)(a^2+3a+6=0.)$

Now,observe that the discriminant of $a^2+3a+6$ is $-15<0,$ there are no real roots.

Therefore $a=3$ is the only root.

$\Rightarrow x+\dfrac{1}{x} =a=3.$

Again, $x^2+\dfrac{1}{x^2}=7,x^3+\dfrac{1}{x^3}=18,x^4+\dfrac{1}{x^4}=47,x^8+\dfrac{1}{x^8}=2207.$

$(x^3+\dfrac{1}{x^3})(x^2+\dfrac{1}{x^2})=x^5+\dfrac{1}{x^5}+(x+\dfrac{1}{x})=(18)(7)=126 \Rightarrow x^5+\dfrac{1}{x^5} = 123.$

$(x^8+\dfrac{1}{x^8})(x^3+\dfrac{1}{x^3})=x^{11}+\dfrac{1}{x^{11}}+(x^5+\dfrac{1}{x^5})=(2207)(18)=39726.\Rightarrow x^{11}+\dfrac{1}{x^{11}} =39726-123 = \boxed{39603}.$

Instead of going in direction of x + (1÷x) and then finding the solution mentioned by chew seong , we can alternatively find x . then put it in the required expression to form a question of simple binomial

If you multiply the bottom equation by x^3, you get x^6-18x^3+1=0. Then if you substitute y=x^3, you get a nice and easy quadratic equation. The solutions for y are 9+-4sqrt(5), which you can substitute the cube root into the problem. Both result in the same answer. To see why these are the only real solutions, factor the equation x^6-18x^3+1=0 (solving this is the same as solving the equation in the problem). let theu two known roots be a and b. Then (x^3-a)(x^3-b)=0 we can then use the identity m^3-n^3=(m-n)(m^2+mn+n^2).. The quadratic here has Nonreal solutions. Now it's clear that all the other roots are not real.