If $x$ is a positive real number such that $x^8 + \frac{1}{x^8} = 47$, then the value of $x^9 + \frac{1}{x^9}$ is

Given: $x^8 + \frac{1}{x^8} = 47$

Find: $x^9 + \frac{1}{x^9}$

---

Let $a = x^4$. This means:

$x^8 = (x^4)^2 = a^2$

$\frac{1}{x^8} = \frac{1}{a^2}$

So our given equation becomes: $a^2 + \frac{1}{a^2} = 47$

---

We need to find $a + \frac{1}{a} = x^4 + \frac{1}{x^4}$.

Using the identity $(a + \frac{1}{a})^2 = a^2 + \frac{1}{a^2} + 2$:

$(a + \frac{1}{a})^2 = 47 + 2 = 49$

$a + \frac{1}{a} = \pm 7$

Since $x$ is positive, $a = x^4$ is also positive, so $a + \frac{1}{a} > 0$.

Therefore: $x^4 + \frac{1}{x^4} = 7$

---

Let $b = x^2$. Then:

$x^4 = (x^2)^2 = b^2$

$\frac{1}{x^4} = \frac{1}{b^2}$

So: $b^2 + \frac{1}{b^2} = 7$

Using the same identity:

$(b + \frac{1}{b})^2 = b^2 + \frac{1}{b^2} + 2 = 7 + 2 = 9$

$x^2 + \frac{1}{x^2} = 3$

---

$(x + \frac{1}{x})^2 = x^2 + \frac{1}{x^2} + 2 = 3 + 2 = 5$

$x + \frac{1}{x} = \sqrt{5}$

---

Using the cubic identity $(x + \frac{1}{x})^3 = x^3 + \frac{1}{x^3} + 3(x + \frac{1}{x})$:

$(\sqrt{5})^3 = x^3 + \frac{1}{x^3} + 3\sqrt{5}$

$5\sqrt{5} = x^3 + \frac{1}{x^3} + 3\sqrt{5}$

$x^3 + \frac{1}{x^3} = 5\sqrt{5} - 3\sqrt{5} = 2\sqrt{5}$

---

Let $d = x^3$, so $x^9 = d^3$ and $\frac{1}{x^9} = \frac{1}{d^3}$

We know: $d + \frac{1}{d} = x^3 + \frac{1}{x^3} = 2\sqrt{5}$

Using the cubic identity again:

$(d + \frac{1}{d})^3 = d^3 + \frac{1}{d^3} + 3(d + \frac{1}{d})$

$(2\sqrt{5})^3 = x^9 + \frac{1}{x^9} + 3(2\sqrt{5})$

$8 \cdot 5\sqrt{5} = x^9 + \frac{1}{x^9} + 6\sqrt{5}$

$40\sqrt{5} = x^9 + \frac{1}{x^9} + 6\sqrt{5}$

$x^9 + \frac{1}{x^9} = 40\sqrt{5} - 6\sqrt{5} = 34\sqrt{5}$

---

This problem demonstrates the power of working with expressions of the form $a + \frac{1}{a}$ and using algebraic identities to connect different powers. The technique of stepping down from higher powers to lower ones then building back up is a common strategy in competitive mathematics.