How to use $x + \frac{1}{x} = 7$ to compute $x^2 + \frac{1}{x^2}$. I am not sure how to approach this question:

You know that $x + \frac{1}{x} = 7$. Compute $x^2 + \frac{1}{x^2}$.

I have tried adding $x + \frac{1}{x}$ to get $\frac{x^2 +1}{x}$ but can't see if this was useful or not. 
I need help in getting started.
 A: Observe that
$$x^2+\frac{1}{x^2}=\left(x+\frac{1}{x}\right
)^2-2.$$
Hope this helps. :)
A: Thank you to Yiyuan Lee and awllower who really answered the question. 
I am only adding this for completeness, and to remind myself of the solution.
We are given $x + \frac{1}{x} =7$ and want to use this to find $x^2 + \frac{1}{x^2}$. 
Using the fact $$\left(x +\frac{1}{x}\right)^2 = x^2 +\frac{1}{x^2} +2$$
and rewriting this as: 
$$\left(x +\frac{1}{x}\right)^2 -2 = x^2 +\frac{1}{x^2}$$
We can substitute $x + \frac{1}{x} =7$ to get:
$$(7)^2 -2 = x^2 +\frac{1}{x^2}$$
So $$x^2 +\frac{1}{x^2} = 47$$
The same method can be used to calculate $x^3 + \frac{1}{x^3}$:
$$\left(x +\frac{1}{x}\right)^3 = x^3 +\frac{1}{x^3} +3x +\frac{3}{x}$$
Noticing that $3x +\frac{3}{x}$ = $3(x+ \frac{1}{x})$, which means $3x +\frac{3}{x} = 3.(7) = 21$
Then it is possible to write:
$$\left(x +\frac{1}{x}\right)^3 -21 = x^3 + \frac{1}{x^3}$$
$$(7)^3 -21 = x^3 + \frac{1}{x^3}$$
$$x^3 + \frac{1}{x^3} = 343 -21$$
$$x^3 + \frac{1}{x^3} = 322$$
A: Notice $\left(x + \frac{1}{x}\right)^2 = x^2 + \frac{1}{x^2} + 2$.
