Solving for \(x^4 + \frac{1}{x^4}\) from \(x + \frac{1}{x}\)

We are given the value of \( \left(x+\frac{1}{x}\right) \) and asked to find the value of \( \left(x^4+\frac{1}{x^4}\right) \). To reach the fourth power from the first power, we can square the expression twice.

Let's perform the calculations step-by-step.

Step-by-Step Calculation to find \(x^4 + \frac{1}{x^4}\)

Step 1: Find the value of \( \left(x^2+\frac{1}{x^2}\right) \)

We are given:

\( \left(x+\frac{1}{x}\right) = 10 \)

Square both sides of the equation:

\( \left(x+\frac{1}{x}\right)^2 = 10^2 \)

Using the algebraic identity \( (a+b)^2 = a^2 + 2ab + b^2 \), where \( a=x \) and \( b=\frac{1}{x} \), we expand the left side:

\( x^2 + 2 \cdot x \cdot \frac{1}{x} + \left(\frac{1}{x}\right)^2 = 100 \)

Simplify the middle term \( 2 \cdot x \cdot \frac{1}{x} = 2 \):

\( x^2 + 2 + \frac{1}{x^2} = 100 \)

Subtract 2 from both sides to isolate \( \left(x^2+\frac{1}{x^2}\right) \):

\( x^2 + \frac{1}{x^2} = 100 - 2 \)

\( \left(x^2+\frac{1}{x^2}\right) = 98 \)

Step 2: Find the value of \( \left(x^4+\frac{1}{x^4}\right) \)

Now that we have the value of \( \left(x^2+\frac{1}{x^2}\right) \), we can square it again to find \( \left(x^4+\frac{1}{x^4}\right) \).

We have:

\( \left(x^2+\frac{1}{x^2}\right) = 98 \)

Square both sides of this equation:

\( \left(x^2+\frac{1}{x^2}\right)^2 = 98^2 \)

Using the same algebraic identity \( (a+b)^2 = a^2 + 2ab + b^2 \), but this time with \( a=x^2 \) and \( b=\frac{1}{x^2} \), we expand the left side:

\( (x^2)^2 + 2 \cdot x^2 \cdot \frac{1}{x^2} + \left(\frac{1}{x^2}\right)^2 = 98^2 \)

Simplify the terms:

\( x^4 + 2 + \frac{1}{x^4} = 98^2 \)

Calculate \( 98^2 \):

\( 98^2 = 98 \times 98 = 9604 \)

Substitute the value of \( 98^2 \) back into the equation:

\( x^4 + 2 + \frac{1}{x^4} = 9604 \)

Subtract 2 from both sides to isolate \( \left(x^4+\frac{1}{x^4}\right) \):

\( x^4 + \frac{1}{x^4} = 9604 - 2 \)

\( \left(x^4+\frac{1}{x^4}\right) = 9602 \)

Thus, the value of \( \left(x^4+\frac{1}{x^4}\right) \) is 9602.

Understanding Related Algebraic Identities

The problem relies heavily on the fundamental algebraic identity for squaring a sum:

• \( (a+b)^2 = a^2 + 2ab + b^2 \)

This identity allows us to easily find the value of \( a^2 + b^2 \) if we know \( a+b \) and \( ab \). In our case, with terms like \( x \) and \( \frac{1}{x} \), the product \( ab = x \cdot \frac{1}{x} \) simplifies nicely to 1, which makes these types of problems straightforward.

Revision Table: Key Algebraic Formulas

Additional Information on Powers of x + 1/x

Problems involving finding values of \( x^n + \frac{1}{x^n} \) given \( x + \frac{1}{x} \) are common. We saw how to go from power 1 to power 2, and then power 2 to power 4 by squaring. You could continue this process to find \( x^8 + \frac{1}{x^8} \), \( x^{16} + \frac{1}{x^{16}} \), and so on.

To find other powers, like \( x^3 + \frac{1}{x^3} \) or \( x^5 + \frac{1}{x^5} \), different identities or combinations of identities are used. For example, to find \( x^3 + \frac{1}{x^3} \), you could cube the original expression \( \left(x+\frac{1}{x}\right) \).

Remembering the core algebraic identities and understanding how the \( 2ab \) term simplifies when \( b = 1/a \) is key to solving these types of algebraic manipulation problems efficiently.