Finding \(x^3 + \frac{1}{x^3}\) when \(x + \frac{1}{x} = 2\)

This problem involves finding the value of an algebraic expression involving powers of x, given a relationship between x and its reciprocal. We are given that \(x + \frac{1}{x} = 2\) and we need to determine the value of \(x^3 + \frac{1}{x^3}\).

Method 1: Using Algebraic Identity

We can use the algebraic identity for the cube of a sum: \((a+b)^3 = a^3 + b^3 + 3ab(a+b)\). Let's consider \(a = x\) and \(b = \frac{1}{x}\).

Cubing the given equation \(x + \frac{1}{x} = 2\), we get:

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

Applying the algebraic identity on the left side:

$$ x^3 + \left(\frac{1}{x}\right)^3 + 3 \cdot x \cdot \frac{1}{x} \left(x + \frac{1}{x}\right) = 8 $$

This simplifies because \(x \cdot \frac{1}{x} = 1\):

$$ x^3 + \frac{1}{x^3} + 3 \cdot 1 \cdot \left(x + \frac{1}{x}\right) = 8 $$

So, we have:

$$ x^3 + \frac{1}{x^3} + 3 \left(x + \frac{1}{x}\right) = 8 $$

We know from the problem statement that \(x + \frac{1}{x} = 2\). Substitute this value into the equation:

$$ x^3 + \frac{1}{x^3} + 3 (2) = 8 $$

$$ x^3 + \frac{1}{x^3} + 6 = 8 $$

Now, isolate \(x^3 + \frac{1}{x^3}\):

$$ x^3 + \frac{1}{x^3} = 8 - 6 $$

$$ x^3 + \frac{1}{x^3} = 2 $$

Thus, the value of \(x^3 + \frac{1}{x^3}\) is 2.

Method 2: Solving for x

We can also solve the given equation \(x + \frac{1}{x} = 2\) to find the value of x directly.

Multiply the entire equation by x to eliminate the fraction:

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

$$ x^2 + 1 = 2x $$

Rearrange the terms to form a quadratic equation:

$$ x^2 - 2x + 1 = 0 $$

This is a perfect square trinomial, which can be factored as:

$$ (x - 1)^2 = 0 $$

Taking the square root of both sides:

$$ x - 1 = 0 $$

Solving for x:

$$ x = 1 $$

Now that we have found \(x = 1\), we can substitute this value into the expression \(x^3 + \frac{1}{x^3}\):

$$ x^3 + \frac{1}{x^3} = 1^3 + \frac{1}{1^3} $$

$$ x^3 + \frac{1}{x^3} = 1 + \frac{1}{1} $$

$$ x^3 + \frac{1}{x^3} = 1 + 1 $$

$$ x^3 + \frac{1}{x^3} = 2 $$

Both methods yield the same result, which is 2.

Summary of the Solution

Given the equation \(x + \frac{1}{x} = 2\), we successfully found the value of \(x^3 + \frac{1}{x^3}\) to be 2 using two different algebraic approaches. The first method utilized the cubic identity \((a+b)^3\) and substitution, while the second method involved solving the quadratic equation to find x and then substituting its value.

Revision Table: Key Algebraic Concepts

Additional Information: Solving \(x + \frac{1}{x} = k\) Problems

Problems involving expressions like \(x + \frac{1}{x}\), \(x^2 + \frac{1}{x^2}\), \(x^3 + \frac{1}{x^3}\), etc., are common in algebra. Here are some useful relationships derived from \(x + \frac{1}{x} = k\):

• To find \(x^2 + \frac{1}{x^2}\): Square the given equation. $$ \left(x + \frac{1}{x}\right)^2 = k^2 $$ $$ x^2 + \left(\frac{1}{x}\right)^2 + 2 \cdot x \cdot \frac{1}{x} = k^2 $$ $$ x^2 + \frac{1}{x^2} + 2 = k^2 $$ $$ x^2 + \frac{1}{x^2} = k^2 - 2 $$
• To find \(x^3 + \frac{1}{x^3}\): Use the identity \(x^3 + \frac{1}{x^3} = \left(x + \frac{1}{x}\right)^3 - 3\left(x + \frac{1}{x}\right)\). $$ x^3 + \frac{1}{x^3} = k^3 - 3k $$

These formulas can save time when solving related algebraic problems. In our specific case, \(k=2\), so \(x^3 + \frac{1}{x^3} = 2^3 - 3(2) = 8 - 6 = 2\).