To find the value of [tex]\( x^3 + \frac{1}{x^3} \)[/tex] given that [tex]\( x + \frac{1}{x} = 7 \)[/tex], we can follow these steps:
1. Express [tex]\( x^2 + \frac{1}{x^2} \)[/tex] in terms of [tex]\( x + \frac{1}{x} \)[/tex]:
We start by squaring both sides of the equation [tex]\( x + \frac{1}{x} = 7 \)[/tex]:
[tex]\[
\left( x + \frac{1}{x} \right)^2 = 7^2
\][/tex]
Expanding the left-hand side, we have:
[tex]\[
x^2 + 2 + \frac{1}{x^2} = 49
\][/tex]
Subtracting 2 from both sides gives:
[tex]\[
x^2 + \frac{1}{x^2} = 49 - 2 = 47
\][/tex]
2. Express [tex]\( x^3 + \frac{1}{x^3} \)[/tex] in terms of [tex]\( x + \frac{1}{x} \)[/tex] and [tex]\( x^2 + \frac{1}{x^2} \)[/tex]:
We use the identity involving cubes:
[tex]\[
(x + \frac{1}{x})(x^2 - 1 + \frac{1}{x^2}) = x^3 + \frac{1}{x^3} + x + \frac{1}{x}
\][/tex]
Plugging in the known values:
[tex]\[
x + \frac{1}{x} = 7 \quad \text{and} \quad x^2 + \frac{1}{x^2} = 47
\][/tex]
Therefore:
[tex]\[
x^2 - 1 + \frac{1}{x^2} = 47 - 1 = 46
\][/tex]
3. Substitute the known values into the identity:
We substitute [tex]\( x + \frac{1}{x} = 7 \)[/tex] and [tex]\( x^2 - 1 + \frac{1}{x^2} = 46 \)[/tex] into the identity:
[tex]\[
(x + \frac{1}{x}) \cdot (x^2 - 1 + \frac{1}{x^2}) = x^3 + \frac{1}{x^3} + x + \frac{1}{x}
\][/tex]
This simplifies to:
[tex]\[
7 \cdot 46 = x^3 + \frac{1}{x^3} + 7
\][/tex]
4. Solve for [tex]\( x^3 + \frac{1}{x^3} \)[/tex]:
To isolate [tex]\( x^3 + \frac{1}{x^3} \)[/tex], subtract 7 from both sides:
[tex]\[
x^3 + \frac{1}{x^3} = 7 \cdot 46 - 7 = 322
\][/tex]
So, the value of [tex]\( x^3 + \frac{1}{x^3} \)[/tex] is [tex]\( \boxed{322} \)[/tex].
