Solving Algebraic Equations to Find Expression Values

The problem asks us to find the value of an algebraic expression, specifically \(x^3 - \frac 1 {x^3}\), given an initial equation involving \(x\). We need to manipulate the given equation to find a useful relationship, likely involving \(x\) and \(\frac{1}{x}\), which can then be used to calculate the desired expression.

Step 1: Simplify the Given Equation

The given equation is \(x\left(3 - \frac 2 x\right) = \frac 3 x\). Let's simplify this equation:

Multiply the term \(x\) inside the parenthesis on the left side:

\(x \times 3 - x \times \frac 2 x = \frac 3 x\)

\(3x - 2 = \frac 3 x\)

Now, rearrange the terms to group the \(x\) and \(\frac{1}{x}\) terms together:

\(3x - \frac 3 x = 2\)

Factor out the common factor, 3, from the terms on the left side:

\(3\left(x - \frac 1 x\right) = 2\)

Divide both sides by 3 to isolate the term \(\left(x - \frac 1 x\right)\):

\(x - \frac 1 x = \frac 2 3\)

This gives us a crucial relationship between \(x\) and \(\frac{1}{x}\).

Step 2: Relate \(x^3 - \frac{1}{x^3}\) to \(x - \frac{1}{x}\)

We need to find the value of \(x^3 - \frac 1 {x^3}\). We can use the algebraic identity for the difference of cubes, which is \(a^3 - b^3 = (a-b)(a^2 + ab + b^2)\). In our case, \(a = x\) and \(b = \frac 1 x\).

So, \(x^3 - \left(\frac 1 x\right)^3 = \left(x - \frac 1 x\right)\left(x^2 + x \cdot \frac 1 x + \left(\frac 1 x\right)^2\right)\)

Simplify the expression inside the second parenthesis:

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

To calculate this, we already know \(x - \frac 1 x = \frac 2 3\). We now need to find the value of \(x^2 + \frac 1 {x^2}\).

Step 3: Find the Value of \(x^2 + \frac{1}{x^2}\)

We know the relationship between \(\left(x - \frac 1 x\right)^2\) and \(x^2 + \frac 1 {x^2}\). Squaring the term \(\left(x - \frac 1 x\right)\):

\(\left(x - \frac 1 x\right)^2 = x^2 - 2 \cdot x \cdot \frac 1 x + \left(\frac 1 x\right)^2\)

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

Rearranging this, we get:

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

Substitute the value of \(x - \frac 1 x = \frac 2 3\):

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

\(x^2 + \frac 1 {x^2} = \frac 4 9 + 2\)

To add \(\frac 4 9\) and 2, find a common denominator:

\(x^2 + \frac 1 {x^2} = \frac 4 9 + \frac {18} 9\)

\(x^2 + \frac 1 {x^2} = \frac {4 + 18} 9 = \frac {22} 9\)

So, \(x^2 + \frac 1 {x^2} = \frac {22} 9\).

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

Now substitute the values of \(\left(x - \frac 1 x\right)\) and \(\left(x^2 + \frac 1 {x^2}\right)\) back into the identity for \(x^3 - \frac 1 {x^3}\):

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

Substitute \(\left(x - \frac 1 x\right) = \frac 2 3\) and \(\left(x^2 + \frac 1 {x^2}\right) = \frac {22} 9\):

\(x^3 - \frac 1 {x^3} = \left(\frac 2 3\right)\left(\frac {22} 9 + 1\right)\)

Add the terms inside the second parenthesis by finding a common denominator:

\(x^3 - \frac 1 {x^3} = \left(\frac 2 3\right)\left(\frac {22} 9 + \frac 9 9\right)\)

\(x^3 - \frac 1 {x^3} = \left(\frac 2 3\right)\left(\frac {22 + 9} 9\right)\)

\(x^3 - \frac 1 {x^3} = \left(\frac 2 3\right)\left(\frac {31} 9\right)\)

Multiply the fractions:

\(x^3 - \frac 1 {x^3} = \frac {2 \times 31} {3 \times 9}\)

\(x^3 - \frac 1 {x^3} = \frac {62} {27}\)

Thus, the value of \(x^3 - \frac 1 {x^3}\) is \(\frac{62}{27}\).

Summary of Calculations

Final Answer Derivation

Starting with the given equation \(x\left(3 - \frac 2 x\right) = \frac 3 x\), we successfully simplified it to find \(x - \frac 1 x = \frac 2 3\). Using the algebraic identity for \(a^3 - b^3\) and the derived value of \(x^2 + \frac 1 {x^2}\), we calculated the required expression \(x^3 - \frac 1 {x^3}\) as \(\frac {62} {27}\).

Revision Table: Algebraic Identities

Additional Information: Solving Equations with Reciprocal Terms

Problems involving expressions like \(x \pm \frac{1}{x}\) and \(x^n \pm \frac{1}{x^n}\) are common in algebra. The key is often to find the value of \(x + \frac{1}{x}\) or \(x - \frac{1}{x}\) first, by simplifying the given equation. Once you have one of these base values, you can find higher powers using identities:

• If you have \(x + \frac{1}{x}\), you can find \(x^2 + \frac{1}{x^2}\) using \(\left(x + \frac{1}{x}\right)^2 = x^2 + 2 + \frac{1}{x^2}\).
• If you have \(x - \frac{1}{x}\), you can find \(x^2 + \frac{1}{x^2}\) using \(\left(x - \frac{1}{x}\right)^2 = x^2 - 2 + \frac{1}{x^2}\).
• To find \(x^3 + \frac{1}{x^3}\) from \(x + \frac{1}{x}\), use \(a^3 + b^3 = (a+b)(a^2 - ab + b^2)\) where \(a=x, b=\frac{1}{x}\). This gives \(x^3 + \frac{1}{x^3} = \left(x + \frac{1}{x}\right)\left(x^2 - 1 + \frac{1}{x^2}\right)\).
• To find \(x^3 - \frac{1}{x^3}\) from \(x - \frac{1}{x}\), use \(a^3 - b^3 = (a-b)(a^2 + ab + b^2)\) where \(a=x, b=\frac{1}{x}\). This gives \(x^3 - \frac{1}{x^3} = \left(x - \frac{1}{x}\right)\left(x^2 + 1 + \frac{1}{x^2}\right)\).

These identities and techniques are fundamental for solving such problems efficiently.