If we are given that $x + \frac{ 1}{x} = a$ , then we know that $x^2 + \frac{ 1}{x^2} = a^2 - 2$ and $x^3 + \frac{ 1}{ x^3} = a^3 - 3a$ . However, it's not too clear how we can relate the latter 2 equations with each other.

Let $x$ be a real number such that $x^3 + \frac{1}{x^3} = 10 \sqrt{2}$ , what is the value of $x^2 + \frac{ 1}{x^2} ?$

The answer is 6.00.

**
This section requires Javascript.
**

You are seeing this because something didn't load right. We suggest you, (a) try
refreshing the page, (b) enabling javascript if it is disabled on your browser and,
finally, (c)
loading the
non-javascript version of this page
. We're sorry about the hassle.

Since $x^3 + \frac{1}{x^3} = 10 \sqrt{2}$ hence $x^ 6 + \frac{ 1}{x^6} = ( 10 \sqrt{2} ) ^2 - 2 = 198$ .

Let $x^2 + \frac{1}{x^2} = a$ , then we have $a^3 - 3a = 198$ . This gives us $(a-6)(a^2 + 6a + 33 ) = 0$ . We have a real root $a = 6$ , and 2 complex roots.