Answer :
Sure, let's solve the given problem step by step.
We start with the equation:
[tex]\[ x + \frac{1}{x} = \sqrt{29} \][/tex]
We need to find the value of [tex]\( x^3 - \frac{1}{x^3} \)[/tex].
First, let's denote:
[tex]\[ y = x + \frac{1}{x} \][/tex]
So, [tex]\( y = \sqrt{29} \)[/tex].
We know an identity that relates [tex]\( y \)[/tex] to [tex]\( x^3 - \frac{1}{x^3} \)[/tex]:
[tex]\[ x^3 - \frac{1}{x^3} = (x + \frac{1}{x}) \left( x^2 - 1 + \frac{1}{x^2} \right) \][/tex]
Thus, we need to find [tex]\( x^2 + \frac{1}{x^2} \)[/tex] first. We can obtain this by squaring both sides of [tex]\( y = x + \frac{1}{x} \)[/tex]:
[tex]\[ \left( x + \frac{1}{x} \right)^2 = y^2 \][/tex]
[tex]\[ x^2 + 2 + \frac{1}{x^2} = y^2 \][/tex]
Substituting [tex]\( y = \sqrt{29} \)[/tex]:
[tex]\[ x^2 + 2 + \frac{1}{x^2} = 29 \][/tex]
[tex]\[ x^2 + \frac{1}{x^2} = 27 \][/tex]
Now we have:
[tex]\[ x^2 + \frac{1}{x^2} = 27 \][/tex]
We can substitute this back into our identity for [tex]\( x^3 - \frac{1}{x^3} \)[/tex]:
[tex]\[ x^3 - \frac{1}{x^3} = (x + \frac{1}{x}) \left( x^2 - 1 + \frac{1}{x^2} \right) \][/tex]
[tex]\[ x^3 - \frac{1}{x^3} = y \left( x^2 + \frac{1}{x^2} - 1 \right) \][/tex]
Since [tex]\( x^2 + \frac{1}{x^2} = 27 \)[/tex], we get:
[tex]\[ x^3 - \frac{1}{x^3} = y \left( 27 - 1 \right) \][/tex]
[tex]\[ x^3 - \frac{1}{x^3} = y \cdot 26 \][/tex]
Now, substitute [tex]\( y = \sqrt{29} \)[/tex] into the equation:
[tex]\[ x^3 - \frac{1}{x^3} = \sqrt{29} \cdot 26 \][/tex]
We know that:
[tex]\[ \sqrt{29} \cdot 26 = 140.014284985497 \][/tex]
Therefore, the value of [tex]\( x^3 - \frac{1}{x^3} \)[/tex] is approximately:
[tex]\[ x^3 - \frac{1}{x^3} \approx 140 \][/tex]
Thus, the correct answer is:
[tex]\[ \boxed{140} \][/tex]
We start with the equation:
[tex]\[ x + \frac{1}{x} = \sqrt{29} \][/tex]
We need to find the value of [tex]\( x^3 - \frac{1}{x^3} \)[/tex].
First, let's denote:
[tex]\[ y = x + \frac{1}{x} \][/tex]
So, [tex]\( y = \sqrt{29} \)[/tex].
We know an identity that relates [tex]\( y \)[/tex] to [tex]\( x^3 - \frac{1}{x^3} \)[/tex]:
[tex]\[ x^3 - \frac{1}{x^3} = (x + \frac{1}{x}) \left( x^2 - 1 + \frac{1}{x^2} \right) \][/tex]
Thus, we need to find [tex]\( x^2 + \frac{1}{x^2} \)[/tex] first. We can obtain this by squaring both sides of [tex]\( y = x + \frac{1}{x} \)[/tex]:
[tex]\[ \left( x + \frac{1}{x} \right)^2 = y^2 \][/tex]
[tex]\[ x^2 + 2 + \frac{1}{x^2} = y^2 \][/tex]
Substituting [tex]\( y = \sqrt{29} \)[/tex]:
[tex]\[ x^2 + 2 + \frac{1}{x^2} = 29 \][/tex]
[tex]\[ x^2 + \frac{1}{x^2} = 27 \][/tex]
Now we have:
[tex]\[ x^2 + \frac{1}{x^2} = 27 \][/tex]
We can substitute this back into our identity for [tex]\( x^3 - \frac{1}{x^3} \)[/tex]:
[tex]\[ x^3 - \frac{1}{x^3} = (x + \frac{1}{x}) \left( x^2 - 1 + \frac{1}{x^2} \right) \][/tex]
[tex]\[ x^3 - \frac{1}{x^3} = y \left( x^2 + \frac{1}{x^2} - 1 \right) \][/tex]
Since [tex]\( x^2 + \frac{1}{x^2} = 27 \)[/tex], we get:
[tex]\[ x^3 - \frac{1}{x^3} = y \left( 27 - 1 \right) \][/tex]
[tex]\[ x^3 - \frac{1}{x^3} = y \cdot 26 \][/tex]
Now, substitute [tex]\( y = \sqrt{29} \)[/tex] into the equation:
[tex]\[ x^3 - \frac{1}{x^3} = \sqrt{29} \cdot 26 \][/tex]
We know that:
[tex]\[ \sqrt{29} \cdot 26 = 140.014284985497 \][/tex]
Therefore, the value of [tex]\( x^3 - \frac{1}{x^3} \)[/tex] is approximately:
[tex]\[ x^3 - \frac{1}{x^3} \approx 140 \][/tex]
Thus, the correct answer is:
[tex]\[ \boxed{140} \][/tex]
