Understanding the Algebraic Problem

The problem asks us to find the value of the expression \( (x^3 + x^{-3}) \) given the equation \( x^8 - 34x^4 + 1 = 0 \) and the condition \( x > 0 \). The expression \( x^{-3} \) is equivalent to \( \frac{1}{x^3} \), so we need to find the value of \( x^3 + \frac{1}{x^3} \).

Solving the Given Equation

We start with the given equation:

\( x^8 - 34x^4 + 1 = 0 \)

Since we are given that \( x > 0 \), \( x^4 \) will also be greater than 0. We can divide the entire equation by \( x^4 \) to simplify it:

\( \frac{x^8}{x^4} - \frac{34x^4}{x^4} + \frac{1}{x^4} = \frac{0}{x^4} \)

\( x^4 - 34 + \frac{1}{x^4} = 0 \)

Rearranging the terms, we get:

\( x^4 + \frac{1}{x^4} = 34 \)

This is a key relationship we will use.

Finding the Value of \( x^2 + \frac{1}{x^2} \)

We know that \( \left(a+b\right)^2 = a^2 + b^2 + 2ab \). Let's apply this identity with \( a = x^2 \) and \( b = \frac{1}{x^2} \):

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

\( \left(x^2 + \frac{1}{x^2}\right)^2 = x^4 + \frac{1}{x^4} + 2 \)

We already found that \( x^4 + \frac{1}{x^4} = 34 \). Substituting this value:

\( \left(x^2 + \frac{1}{x^2}\right)^2 = 34 + 2 \)

\( \left(x^2 + \frac{1}{x^2}\right)^2 = 36 \)

Taking the square root of both sides:

\( x^2 + \frac{1}{x^2} = \pm\sqrt{36} \)

\( x^2 + \frac{1}{x^2} = \pm 6 \)

Since \( x > 0 \), \( x^2 \) is positive, and \( \frac{1}{x^2} \) is positive. The sum of two positive numbers must be positive. Therefore, we take the positive root:

\( x^2 + \frac{1}{x^2} = 6 \)

This is another important relationship.

Finding the Value of \( x + \frac{1}{x} \)

We use the identity \( \left(a+b\right)^2 = a^2 + b^2 + 2ab \) again, this time with \( a = x \) and \( b = \frac{1}{x} \):

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

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

We just found that \( x^2 + \frac{1}{x^2} = 6 \). Substituting this value:

\( \left(x + \frac{1}{x}\right)^2 = 6 + 2 \)

\( \left(x + \frac{1}{x}\right)^2 = 8 \)

Taking the square root of both sides:

\( x + \frac{1}{x} = \pm\sqrt{8} \)

\( x + \frac{1}{x} = \pm\sqrt{4 \times 2} \)

\( x + \frac{1}{x} = \pm 2\sqrt{2} \)

Since \( x > 0 \), \( x \) is positive, and \( \frac{1}{x} \) is positive. The sum of two positive numbers must be positive. Therefore, we take the positive root:

\( x + \frac{1}{x} = 2\sqrt{2} \)

This is the final relationship we need.

Calculating the Value of \( x^3 + x^{-3} \)

We need to find the value of \( x^3 + \frac{1}{x^3} \). We can use the identity \( a^3 + b^3 = (a+b)(a^2 - ab + b^2) \). Let \( a = x \) and \( b = \frac{1}{x} \):

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

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

We know \( x + \frac{1}{x} = 2\sqrt{2} \) and \( x^2 + \frac{1}{x^2} = 6 \). Substitute these values into the expression:

\( x^3 + \frac{1}{x^3} = \left(2\sqrt{2}\right) \left(6 - 1\right) \)

\( x^3 + \frac{1}{x^3} = \left(2\sqrt{2}\right) \left(5\right) \)

\( x^3 + \frac{1}{x^3} = 10\sqrt{2} \)

Alternatively, we can use the identity \( \left(a+b\right)^3 = a^3 + b^3 + 3ab(a+b) \). Let \( a=x \) and \( b=\frac{1}{x} \):

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

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

We know \( x + \frac{1}{x} = 2\sqrt{2} \). Substitute this value:

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

Calculate \( \left(2\sqrt{2}\right)^3 \):

\( \left(2\sqrt{2}\right)^3 = 2^3 \times \left(\sqrt{2}\right)^3 = 8 \times \left(\sqrt{2} \times \sqrt{2} \times \sqrt{2}\right) = 8 \times \left(2 \times \sqrt{2}\right) = 8 \times 2\sqrt{2} = 16\sqrt{2} \)

Substitute this back into the equation:

\( 16\sqrt{2} = x^3 + \frac{1}{x^3} + 6\sqrt{2} \)

Solve for \( x^3 + \frac{1}{x^3} \):

\( x^3 + \frac{1}{x^3} = 16\sqrt{2} - 6\sqrt{2} \)

\( x^3 + \frac{1}{x^3} = 10\sqrt{2} \)

Both methods yield the same result.

Summary of Steps

Here is a brief recap of the steps taken to find the value of \( x^3 + x^{-3} \):

1. Start with the given equation \( x^8 - 34x^4 + 1 = 0 \).
2. Divide by \( x^4 \) to get \( x^4 + \frac{1}{x^4} = 34 \).
3. Use the identity \( (a+b)^2 \) to find \( x^2 + \frac{1}{x^2} \) from \( x^4 + \frac{1}{x^4} \).
4. Use the identity \( (a+b)^2 \) again to find \( x + \frac{1}{x} \) from \( x^2 + \frac{1}{x^2} \).
5. Use the identity for \( a^3 + b^3 \) or \( (a+b)^3 \) to find \( x^3 + \frac{1}{x^3} \) from \( x + \frac{1}{x} \) and \( x^2 + \frac{1}{x^2} \).
6. Ensure positive roots are taken due to the condition \( x > 0 \).

Evaluation of Options

The value we calculated for \( (x^3 + x^{-3}) \) is \( 10\sqrt{2} \). Let's compare this with the given options:

• Option 1: \( 6\sqrt{2} \)
• Option 2: \( 5\sqrt{4} = 5 \times 2 = 10 \)
• Option 3: \( 10\sqrt{2} \)
• Option 4: \( 5\sqrt{6} \)

Our calculated value matches Option 3.

Revision Table: Key Identities Used

Additional Information: Working with Negative Exponents

Understanding negative exponents is crucial for solving this problem. A term with a negative exponent means the reciprocal of the base raised to the positive exponent.

• For example, \( x^{-n} = \frac{1}{x^n} \).
• In our problem, \( x^{-3} = \frac{1}{x^3} \).
• This allows us to rewrite the expression \( x^3 + x^{-3} \) as \( x^3 + \frac{1}{x^3} \).
• This transformation is essential for applying the algebraic identities for sums of powers.

Also, remember that for any non-zero number \( a \), \( a^0 = 1 \). In our derivation, when we have terms like \( x \cdot \frac{1}{x} \), they simplify to \( x^{1} \cdot x^{-1} = x^{1-1} = x^0 = 1 \).