If x^2 - 9x + 1 = 0, then what is the value of x^3 + 1/x^3?

Steps to Solve

1. Solve the Quadratic Equation

First, we need to find the roots of the quadratic equation $x^2 - 9x + 1 = 0$ using the quadratic formula:

$$ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} $$

Here, ( a = 1 ), ( b = -9 ), and ( c = 1 ). Plugging these values into the formula:

$$ x = \frac{9 \pm \sqrt{(-9)^2 - 4 \cdot 1 \cdot 1}}{2 \cdot 1} $$

Calculate the discriminant:

$$ (-9)^2 - 4 \cdot 1 \cdot 1 = 81 - 4 = 77 $$

Thus,

$$ x = \frac{9 \pm \sqrt{77}}{2} $$

2. Calculate $x + \frac{1}{x}$

To find $x^3 + \frac{1}{x^3}$, we first calculate $x + \frac{1}{x}$.

Using the roots from the quadratic formula:

Let ( x = \frac{9 + \sqrt{77}}{2} ), then:

$$ \frac{1}{x} = \frac{2}{9 + \sqrt{77}} $$

Rationalizing the denominator leads us to:

$$ x + \frac{1}{x} = \frac{9 + \sqrt{77}}{2} + \frac{2(9 - \sqrt{77})}{(9 + \sqrt{77})(9 - \sqrt{77})} $$

We can simplify this further, but it's easier to jump to finding $x^2 + \frac{1}{x^2}$.

3. Calculate $x^2 + \frac{1}{x^2}$

Using the identity:

$$ x^2 + \frac{1}{x^2} = \left( x + \frac{1}{x} \right)^2 - 2 $$

Let ( t = x + \frac{1}{x} ).

We will substitute values but let's be aware of calculating ( t ):

Using a direct result from $x - 9 + \frac{1}{x} = -8$, we find ( t = 9 ).

Thus,

$$ x^2 + \frac{1}{x^2} = 9^2 - 2 = 81 - 2 = 79 $$

4. Calculate $x^3 + \frac{1}{x^3}$

Using the identity:

$$ x^3 + \frac{1}{x^3} = \left( x + \frac{1}{x} \right) \left( x^2 + \frac{1}{x^2} \right) - \left( x + \frac{1}{x} \right) $$

Substituting values:

$$ x^3 + \frac{1}{x^3} = t \cdot (79) - t $$

$$ = 9 \cdot 79 - 9 $$

5. Final calculation

Calculating gives us:

$$ 9 \cdot 79 - 9 = 711 - 9 = 702 $$