If 2x – y = 2 and xy = \(\frac{3}{2}\) , then what is the value of x 3– \(\frac{{{y^3}}}{8}\) ?
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:
- \(\small 2x - y = 2\)
- \(\small xy = \frac{3}{2}\)
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
| Concept | Formula / Identity | Application in Problem |
|---|---|---|
| Difference of Cubes | \(\small a^3 - b^3 = (a - b)(a^2 + ab + b^2)\) | Used to expand \(\small x^3 - \frac{y^3}{8}\) |
| Squaring a Binomial | \(\small (a - b)^2 = a^2 - 2ab + b^2\) | Used to relate \(\small (x - \frac{y}{2})^2\) to \(\small x^2 + \frac{y^2}{4}\) and \(\small xy\) |
| Solving Linear Equations | Basic algebraic operations (division, addition) | Used to derive \(\small x - \frac{y}{2}\) from \(\small 2x - y = 2\) |
| Substituting Values | Replacing variables with known values | Used to substitute \(\small xy\) value and factor values |
| Fraction Arithmetic | Adding fractions with different denominators | Used to calculate sums like \(\small 1 + \frac{3}{2}\) and \(\small \frac{5}{2} + \frac{3}{4}\) |
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.
- From \(\small 2x - y = 2\), express \(\small y\) in terms of \(\small x\): \(\small y = 2x - 2\).
- Substitute this into the second equation \(\small xy = \frac{3}{2}\): \(\small x(2x - 2) = \frac{3}{2}\).
- This leads to a quadratic equation in \(\small x\): \(\small 2x^2 - 2x = \frac{3}{2}\), which simplifies to \(\small 4x^2 - 4x - 3 = 0\).
- Solve the quadratic equation for \(\small x\). Factoring gives \(\small (2x+1)(2x-3)=0\), yielding \(\small x = -\frac{1}{2}\) or \(\small x = \frac{3}{2}\).
- 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 \(\small xy = (\frac{3}{2})(1) = \frac{3}{2}\).
- 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.
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