Understanding the Algebra Problem

This problem gives us two equations relating two variables, \(x\) and \(y\), and asks us to find the value of a specific algebraic expression.

The given information is:

• Equation 1: \(x - y = 11\)
• Equation 2: \(\frac{1}{x} - \frac{1}{y} = \frac{11}{24}\)

We need to calculate the value of the expression \(x^3 - y^3 + x^2y^2\).

Simplifying Equation 2 and Finding the Product xy

Let's start by simplifying the second equation. The left side involves fractions with \(x\) and \(y\) in the denominators. We can combine these fractions by finding a common denominator, which is \(xy\):

\(\frac{1}{x} - \frac{1}{y} = \frac{y}{xy} - \frac{x}{xy} = \frac{y - x}{xy}\)

So, Equation 2 can be rewritten as:

\(\frac{y - x}{xy} = \frac{11}{24}\)

Now, let's look at Equation 1: \(x - y = 11\). Notice that the numerator in our simplified Equation 2 is \(y - x\), which is the negative of \(x - y\). So, \(y - x = -(x - y) = -11\).

Substitute this value into the simplified Equation 2:

\(\frac{-11}{xy} = \frac{11}{24}\)

To find \(xy\), we can cross-multiply or observe the relationship between the numerators. Since the numerators are \(-11\) and \(11\), the denominators must have a similar relationship. Specifically, if \(\frac{a}{b} = \frac{c}{d}\), then \(ad = bc\).

\(-11 \times 24 = 11 \times xy\)

\(-264 = 11xy\)

Now, divide both sides by 11 to solve for \(xy\):

\(xy = \frac{-264}{11}\)

\(xy = -24\)

So, we have successfully found the value of \(xy\).

Finding \(x^2 + y^2\) Using Algebraic Identities

We have two important pieces of information now: \(x - y = 11\) and \(xy = -24\).

The expression we need to evaluate contains terms like \(x^3\), \(y^3\), and \(x^2y^2\). To work with \(x^3 - y^3\), we can use the difference of cubes formula, which involves \(x-y\) and \(x^2 + xy + y^2\).

We already have \(x - y\) and \(xy\). We need to find \(x^2 + y^2\). We can use the algebraic identity for squaring a difference:

\((x - y)^2 = x^2 - 2xy + y^2\)

We can rearrange this identity to find \(x^2 + y^2\):

\(x^2 + y^2 = (x - y)^2 + 2xy\)

Now, substitute the values we know:

\(x^2 + y^2 = (11)^2 + 2(-24)\)

\(x^2 + y^2 = 121 + (-48)\)

\(x^2 + y^2 = 121 - 48\)

\(x^2 + y^2 = 73\)

So, we have found that \(x^2 + y^2 = 73\).

Evaluating the Target Expression \(x^3 - y^3 + x^2y^2\)

The expression we need to find the value of is \(x^3 - y^3 + x^2y^2\).

Let's break this down into parts:

Part 1: Calculate \(x^3 - y^3\)

We use the difference of cubes formula:

\(x^3 - y^3 = (x - y)(x^2 + xy + y^2)\)

We know \(x - y = 11\) and \(xy = -24\). We also found \(x^2 + y^2 = 73\).

Let's calculate the term \(x^2 + xy + y^2\):

\(x^2 + xy + y^2 = (x^2 + y^2) + xy\)

\(x^2 + xy + y^2 = 73 + (-24)\)

\(x^2 + xy + y^2 = 73 - 24\)

\(x^2 + xy + y^2 = 49\)

Now substitute the values of \(x - y\) and \(x^2 + xy + y^2\) into the \(x^3 - y^3\) formula:

\(x^3 - y^3 = (11)(49)\)

\(x^3 - y^3 = 539\)

Part 2: Calculate \(x^2y^2\)

We know that \(xy = -24\). The term \(x^2y^2\) is the square of \(xy\).

\(x^2y^2 = (xy)^2\)

\(x^2y^2 = (-24)^2\)

When squaring a negative number, the result is positive:

\(x^2y^2 = (-24) \times (-24) = 576\)

Part 3: Combine the results

Now we add the values of \(x^3 - y^3\) and \(x^2y^2\) to find the value of the full expression:

\(x^3 - y^3 + x^2y^2 = (x^3 - y^3) + (x^2y^2)\)

\(x^3 - y^3 + x^2y^2 = 539 + 576\)

Adding these two numbers:

\(539 + 576 = 1115\)

So, the value of the expression \(x^3 - y^3 + x^2y^2\) is 1115.

Summary of Calculations

The calculated value matches one of the given options.

Revision Table: Key Formulas Used in Solving the Problem

Additional Information: Alternative Approaches or Concepts

In this problem, we solved for the expression \(x^3 - y^3 + x^2y^2\) without needing to find the individual values of \(x\) and \(y\). This was possible because the expression could be evaluated directly using the derived values of \(x-y\) and \(xy\).

However, it is also possible to find the values of \(x\) and \(y\). We have the system of equations:

• \(x - y = 11\)
• \(xy = -24\)

Consider a quadratic equation whose roots are \(x\) and \(y\). This equation is given by \(t^2 - (x + y)t + xy = 0\). We know \(xy = -24\). We need to find \(x + y\).

We can use the identity relating the sum, difference, and product of two numbers:

\((x + y)^2 = (x - y)^2 + 4xy\)

Substitute the known values:

\((x + y)^2 = (11)^2 + 4(-24)\)

\((x + y)^2 = 121 - 96\)

\((x + y)^2 = 25\)

Taking the square root of both sides, we get \(x + y = \pm \sqrt{25}\), so \(x + y = 5\) or \(x + y = -5\).

Case 1: \(x + y = 5\) and \(x - y = 11\)

Adding the two equations: \((x + y) + (x - y) = 5 + 11 \implies 2x = 16 \implies x = 8\).

Substituting \(x = 8\) into \(x - y = 11\): \(8 - y = 11 \implies y = 8 - 11 = -3\).

Check \(xy = -24\): \(8 \times (-3) = -24\). This pair \((x, y) = (8, -3)\) works.

Case 2: \(x + y = -5\) and \(x - y = 11\)

Adding the two equations: \((x + y) + (x - y) = -5 + 11 \implies 2x = 6 \implies x = 3\).

Substituting \(x = 3\) into \(x - y = 11\): \(3 - y = 11 \implies y = 3 - 11 = -8\).

Check \(xy = -24\): \(3 \times (-8) = -24\). This pair \((x, y) = (3, -8)\) also works.

Both pairs \((8, -3)\) and \((3, -8)\) satisfy the original equations. If we were to substitute either pair into the expression \(x^3 - y^3 + x^2y^2\), we would get the same result, 1115, confirming our solution.