Calculate \(x^8 - \frac{1}{x^8}\) from \(x + \frac{1}{x}\) Equation

We are given the equation \(x + \frac{1}{x} = 3\sqrt{2}\) and the condition that \(x > 1\). Our goal is to find the value of \(x^8 - \frac{1}{x^8}\).

Understanding the \(x^8 - \frac{1}{x^8}\) Algebraic Expression

The expression \(x^8 - \frac{1}{x^8}\) can be factored using the difference of squares identity, \(a^2 - b^2 = (a+b)(a-b)\). We can apply this identity multiple times:

\[x^8 - \frac{1}{x^8} = \left(x^4\right)^2 - \left(\frac{1}{x^4}\right)^2 = \left(x^4 + \frac{1}{x^4}\right)\left(x^4 - \frac{1}{x^4}\right)\] \[x^4 - \frac{1}{x^4} = \left(x^2\right)^2 - \left(\frac{1}{x^2}\right)^2 = \left(x^2 + \frac{1}{x^2}\right)\left(x^2 - \frac{1}{x^2}\right)\] \[x^2 - \frac{1}{x^2} = \left(x\right)^2 - \left(\frac{1}{x}\right)^2 = \left(x + \frac{1}{x}\right)\left(x - \frac{1}{x}\right)\]

Combining these, we get:

\[x^8 - \frac{1}{x^8} = \left(x^4 + \frac{1}{x^4}\right)\left(x^2 + \frac{1}{x^2}\right)\left(x + \frac{1}{x}\right)\left(x - \frac{1}{x}\right)\]

To find the value of \(x^8 - \frac{1}{x^8}\), we need to calculate the values of \(x - \frac{1}{x}\), \(x^2 + \frac{1}{x^2}\), and \(x^4 + \frac{1}{x^4}\) using the given \(x + \frac{1}{x}\).

Step-by-Step Calculation for Finding the Value

Step 1: Calculate \(x^2 + \frac{1}{x^2}\)

We use the identity \(\left(a+b\right)^2 = a^2 + b^2 + 2ab\). Here, \(a=x\) and \(b=\frac{1}{x}\).

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

Rearranging the formula, we get \(x^2 + \frac{1}{x^2} = \left(x + \frac{1}{x}\right)^2 - 2\). Given \(x + \frac{1}{x} = 3\sqrt{2}\), we substitute this value:

\[x^2 + \frac{1}{x^2} = \left(3\sqrt{2}\right)^2 - 2\] \[x^2 + \frac{1}{x^2} = (9 \times 2) - 2\] \[x^2 + \frac{1}{x^2} = 18 - 2\] \[x^2 + \frac{1}{x^2} = 16\]

Step 2: Calculate \(x - \frac{1}{x}\)

We use the identity \(\left(a-b\right)^2 = \left(a+b\right)^2 - 4ab\). Here, \(a=x\) and \(b=\frac{1}{x}\).

\[\left(x - \frac{1}{x}\right)^2 = \left(x + \frac{1}{x}\right)^2 - 4 \cdot x \cdot \frac{1}{x}\] \[\left(x - \frac{1}{x}\right)^2 = \left(x + \frac{1}{x}\right)^2 - 4\]

Given \(x + \frac{1}{x} = 3\sqrt{2}\), we substitute this value:

\[\left(x - \frac{1}{x}\right)^2 = \left(3\sqrt{2}\right)^2 - 4\] \[\left(x - \frac{1}{x}\right)^2 = 18 - 4\] \[\left(x - \frac{1}{x}\right)^2 = 14\]

Taking the square root of both sides, \(x - \frac{1}{x} = \pm\sqrt{14}\). We are given that \(x > 1\). If \(x > 1\), then \(\frac{1}{x} < 1\), which implies \(x - \frac{1}{x}\) must be positive. Therefore, we take the positive root:

\[x - \frac{1}{x} = \sqrt{14}\]

Step 3: Calculate \(x^4 + \frac{1}{x^4}\)

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

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

Rearranging the formula, we get \(x^4 + \frac{1}{x^4} = \left(x^2 + \frac{1}{x^2}\right)^2 - 2\). We found \(x^2 + \frac{1}{x^2} = 16\) in Step 1. Substitute this value:

\[x^4 + \frac{1}{x^4} = \left(16\right)^2 - 2\] \[x^4 + \frac{1}{x^4} = 256 - 2\] \[x^4 + \frac{1}{x^4} = 254\]

Step 4: Calculate \(x^8 - \frac{1}{x^8}\)

Now substitute the values we found into the factored expression:

\[x^8 - \frac{1}{x^8} = \left(x^4 + \frac{1}{x^4}\right)\left(x^2 + \frac{1}{x^2}\right)\left(x + \frac{1}{x}\right)\left(x - \frac{1}{x}\right)\] \[x^8 - \frac{1}{x^8} = \left(254\right)\left(16\right)\left(3\sqrt{2}\right)\left(\sqrt{14}\right)\]

Multiply the terms:

\[x^8 - \frac{1}{x^8} = (254 \times 16 \times 3) \times (\sqrt{2} \times \sqrt{14})\] \[x^8 - \frac{1}{x^8} = (254 \times 48) \times (\sqrt{2 \times 14})\] \[x^8 - \frac{1}{x^8} = (254 \times 48) \times \sqrt{28}\]

Simplify the square root:

\[\sqrt{28} = \sqrt{4 \times 7} = \sqrt{4} \times \sqrt{7} = 2\sqrt{7}\]

Substitute this back into the expression:

\[x^8 - \frac{1}{x^8} = (254 \times 48) \times 2\sqrt{7}\]

Calculate \(254 \times 48\):

\[254 \times 48 = 12192\]

Substitute this value:

\[x^8 - \frac{1}{x^8} = 12192 \times 2\sqrt{7}\] \[x^8 - \frac{1}{x^8} = 24384\sqrt{7}\]

Key Algebraic Identities Used

• Difference of Squares: \(a^2 - b^2 = (a+b)(a-b)\)
• Square of Sum: \((a+b)^2 = a^2 + b^2 + 2ab\)
• Square of Difference: \((a-b)^2 = a^2 + b^2 - 2ab = (a+b)^2 - 4ab\)

Revision Table: Summary of Calculated Values for the Equation

Additional Information on Algebraic Powers and Equations

Problems involving powers of \(x\) and \(\frac{1}{x}\) often utilize the fundamental algebraic identities related to squares and differences of squares. Recognising the pattern \(a^2 - b^2 = (a+b)(a-b)\) at different powers (like \(x^8 - \frac{1}{x^8}\) as \(\left(x^4\right)^2 - \left(\frac{1}{x^4}\right)^2\)) is key.

The condition \(x > 1\) is crucial when finding \(x - \frac{1}{x}\). Without this condition, we would have \(x - \frac{1}{x} = \pm\sqrt{14}\), and the final answer would have two possible values, \(\pm 24384\sqrt{7}\). The condition \(x > 1\) ensures that \(x\) is larger than \(\frac{1}{x}\), making \(x - \frac{1}{x}\) positive.