Finding the Value of an Algebraic Expression

We are given two equations and asked to find the value of a specific algebraic expression. The given equations are:

1. \(\small 2x - y = 2\)
2. \(\small xy = \frac{3}{2}\)

We need to find the value of the expression \(\small x^3 - \frac{y^3}{8}\).

Analyzing the Expression \(\small x^3 - \frac{y^3}{8}\)

The expression \(\small x^3 - \frac{y^3}{8}\) can be rewritten as \(\small x^3 - \left(\frac{y}{2}\right)^3\). This is in the form of a difference of cubes, \(\small a^3 - b^3\), where \(\small a = x\) and \(\small b = \frac{y}{2}\).

The formula for the difference of cubes is:

\(\small a^3 - b^3 = (a - b)(a^2 + ab + b^2)\)

Applying this formula to our expression:

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

\(\small x^3 - \frac{y^3}{8} = \left(x - \frac{y}{2}\right)\left(x^2 + \frac{xy}{2} + \frac{y^2}{4}\right)\)

Using the Given Equations to Find the Factors

We need to find the values of the two factors: \(\small \left(x - \frac{y}{2}\right)\) and \(\small \left(x^2 + \frac{xy}{2} + \frac{y^2}{4}\right)\).

Finding the Value of \(\small \left(x - \frac{y}{2}\right)\)

From the first given equation, \(\small 2x - y = 2\).

Divide both sides of this equation by 2:

\(\small \frac{2x - y}{2} = \frac{2}{2}\)

\(\small \frac{2x}{2} - \frac{y}{2} = 1\)

\(\small x - \frac{y}{2} = 1\)

So, the value of the first factor is 1.

Finding the Value of \(\small \left(x^2 + \frac{xy}{2} + \frac{y^2}{4}\right)\)

This factor contains the terms \(\small x^2\), \(\small \frac{xy}{2}\), and \(\small \frac{y^2}{4}\).

We are given \(\small xy = \frac{3}{2}\). So, \(\small \frac{xy}{2} = \frac{1}{2} \times \frac{3}{2} = \frac{3}{4}\).

The expression becomes \(\small x^2 + \frac{3}{4} + \frac{y^2}{4}\).

Now we need to find the value of \(\small x^2 + \frac{y^2}{4}\).

Consider the equation \(\small x - \frac{y}{2} = 1\). Square both sides of this equation:

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

Using the formula \(\small (a - b)^2 = a^2 - 2ab + b^2\):

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

\(\small x^2 - xy + \frac{y^2}{4} = 1\)

We know that \(\small xy = \frac{3}{2}\). Substitute this value:

\(\small x^2 - \frac{3}{2} + \frac{y^2}{4} = 1\)

Add \(\small \frac{3}{2}\) to both sides:

\(\small x^2 + \frac{y^2}{4} = 1 + \frac{3}{2}\)

\(\small x^2 + \frac{y^2}{4} = \frac{2}{2} + \frac{3}{2} = \frac{2+3}{2} = \frac{5}{2}\)

Now substitute this value back into the second factor \(\small \left(x^2 + \frac{xy}{2} + \frac{y^2}{4}\right)\):

\(\small x^2 + \frac{xy}{2} + \frac{y^2}{4} = \left(x^2 + \frac{y^2}{4}\right) + \frac{xy}{2} = \frac{5}{2} + \frac{3}{4}\)

To add these fractions, find a common denominator, which is 4:

\(\small \frac{5}{2} + \frac{3}{4} = \frac{5 \times 2}{2 \times 2} + \frac{3}{4} = \frac{10}{4} + \frac{3}{4} = \frac{10+3}{4} = \frac{13}{4}\)

So, the value of the second factor is \(\small \frac{13}{4}\).

Calculating the Final Value

The expression \(\small x^3 - \frac{y^3}{8}\) is equal to the product of the two factors:

\(\small \left(x - \frac{y}{2}\right) \times \left(x^2 + \frac{xy}{2} + \frac{y^2}{4}\right)\)

Substitute the values we found:

\(\small 1 \times \frac{13}{4} = \frac{13}{4}\)

Therefore, the value of \(\small x^3 - \frac{y^3}{8}\) is \(\small \frac{13}{4}\).

Checking the Options

The calculated value \(\small \frac{13}{4}\) matches one of the given options.

The correct option is \(\small \frac{13}{4}\).

Revision Table: Key Algebraic Concepts

Additional Information: Alternative Approach

While the difference of cubes formula provides an elegant solution, another approach involves solving for the values of \(\small x\) and \(\small y\) directly using the given equations.

1. From \(\small 2x - y = 2\), express \(\small y\) in terms of \(\small x\): \(\small y = 2x - 2\).
2. Substitute this into the second equation \(\small xy = \frac{3}{2}\): \(\small x(2x - 2) = \frac{3}{2}\).
3. This leads to a quadratic equation in \(\small x\): \(\small 2x^2 - 2x = \frac{3}{2}\), which simplifies to \(\small 4x^2 - 4x - 3 = 0\).
4. Solve the quadratic equation for \(\small x\). Factoring gives \(\small (2x+1)(2x-3)=0\), yielding \(\small x = -\frac{1}{2}\) or \(\small x = \frac{3}{2}\).
5. For each value of \(\small x\), find the corresponding value of \(\small y\) using \(\small y = 2x - 2\).
   • If \(\small x = -\frac{1}{2}\), \(\small y = 2(-\frac{1}{2}) - 2 = -1 - 2 = -3\). Check \(\small xy = (-\frac{1}{2})(-3) = \frac{3}{2}\).
   • If \(\small x = \frac{3}{2}\), \(\small y = 2(\frac{3}{2}) - 2 = 3 - 2 = 1\). Check \(\small xy = (\frac{3}{2})(1) = \frac{3}{2}\).
6. Calculate \(\small x^3 - \frac{y^3}{8}\) for each pair of (\(\small x, y\)).
   • For \(\small x = -\frac{1}{2}, y = -3\): \(\small (-\frac{1}{2})^3 - \frac{(-3)^3}{8} = -\frac{1}{8} - \frac{-27}{8} = -\frac{1}{8} + \frac{27}{8} = \frac{26}{8} = \frac{13}{4}\).
   • For \(\small x = \frac{3}{2}, y = 1\): \(\small (\frac{3}{2})^3 - \frac{1^3}{8} = \frac{27}{8} - \frac{1}{8} = \frac{26}{8} = \frac{13}{4}\).

Both pairs of solutions for (\(\small x, y\)) yield the same value for the expression, confirming the result obtained using the algebraic identity approach.