Understanding the Problem: Calculating \(x^3 - y^3\)

The question asks us to find the value of \(x^3 - y^3\) given specific expressions for \(x\) and \(y\). Both \(x\) and \(y\) are defined as fractions involving square roots. To solve this, we first need to simplify the expressions for \(x\) and \(y\), and then use the formula for the difference of cubes or other algebraic methods to find \(x^3 - y^3\).

Simplifying \(x\) and \(y\) Using Rationalization

The expressions for \(x\) and \(y\) have denominators with square roots. We can simplify these expressions by rationalizing the denominator. This involves multiplying the numerator and denominator by the conjugate of the denominator.

Step 1: Simplify \(x\)

Given \(x = \frac{\sqrt{3}+1}{\sqrt{3}-1}\). The conjugate of the denominator \((\sqrt{3}-1)\) is \((\sqrt{3}+1)\). Multiply the numerator and denominator by \((\sqrt{3}+1)\):

\(x = \frac{\sqrt{3}+1}{\sqrt{3}-1} \times \frac{\sqrt{3}+1}{\sqrt{3}+1}\)

In the numerator, we have \((\sqrt{3}+1)^2\), which expands to \((\sqrt{3})^2 + 2(\sqrt{3})(1) + 1^2\). In the denominator, we have a difference of squares, \((\sqrt{3}-1)(\sqrt{3}+1)\), which is \((\sqrt{3})^2 - 1^2\).

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

\(x = \frac{3 + 2\sqrt{3} + 1}{3 - 1}\)

\(x = \frac{4 + 2\sqrt{3}}{2}\)

Now, divide both terms in the numerator by 2:

\(x = \frac{4}{2} + \frac{2\sqrt{3}}{2}\)

\(x = 2 + \sqrt{3}\)

Step 2: Simplify \(y\)

Given \(y = \frac{\sqrt{3}-1}{\sqrt{3}+1}\). Notice that \(y\) is the reciprocal of \(x\). Since we have simplified \(x\) to \(2+\sqrt{3}\), we can say \(y = \frac{1}{x} = \frac{1}{2+\sqrt{3}}\).

To rationalize the denominator of \(y\), multiply the numerator and denominator by the conjugate of \((2+\sqrt{3})\), which is \((2-\sqrt{3})\):

\(y = \frac{1}{2+\sqrt{3}} \times \frac{2-\sqrt{3}}{2-\sqrt{3}}\)

In the numerator, we have \(1 \times (2-\sqrt{3}) = 2-\sqrt{3}\). In the denominator, we have a difference of squares, \((2+\sqrt{3})(2-\sqrt{3})\), which is \(2^2 - (\sqrt{3})^2\).

\(y = \frac{2-\sqrt{3}}{2^2 - (\sqrt{3})^2}\)

\(y = \frac{2-\sqrt{3}}{4 - 3}\)

\(y = \frac{2-\sqrt{3}}{1}\)

\(y = 2 - \sqrt{3}\)

So, we have simplified the expressions to \(x = 2 + \sqrt{3}\) and \(y = 2 - \sqrt{3}\).

Calculating \(x-y\) and \(xy\)

Before calculating \(x^3 - y^3\), let's find the values of \(x-y\) and \(xy\), as these are often useful in algebraic formulas.

\(x - y = (2 + \sqrt{3}) - (2 - \sqrt{3})\)

\(x - y = 2 + \sqrt{3} - 2 + \sqrt{3}\)

\(x - y = 2\sqrt{3}\)

\(x \times y = (2 + \sqrt{3})(2 - \sqrt{3})\)

This is a difference of squares formula, \((a+b)(a-b) = a^2 - b^2\):

\(x \times y = 2^2 - (\sqrt{3})^2\)

\(x \times y = 4 - 3\)

\(x \times y = 1\)

Finding \(x^3 - y^3\) Using the Difference of Cubes Formula

We need to find the value of \(x^3 - y^3\). The formula for the difference of cubes is \(a^3 - b^3 = (a-b)(a^2 + ab + b^2)\). Alternatively, a useful form for calculation when \(a-b\) and \(ab\) are known is \(a^3 - b^3 = (a-b)^3 + 3ab(a-b)\).

Using the second formula with \(a=x\) and \(b=y\):

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

Substitute the values we found for \(x-y\) and \(xy\):

\(x-y = 2\sqrt{3}\)

\(xy = 1\)

\(x^3 - y^3 = (2\sqrt{3})^3 + 3(1)(2\sqrt{3})\)

Step 3: Calculate \((2\sqrt{3})^3\)

\((2\sqrt{3})^3 = 2^3 \times (\sqrt{3})^3\)

\(2^3 = 2 \times 2 \times 2 = 8\)

\((\sqrt{3})^3 = \sqrt{3} \times \sqrt{3} \times \sqrt{3} = (\sqrt{3} \times \sqrt{3}) \times \sqrt{3} = 3 \times \sqrt{3} = 3\sqrt{3}\)

So, \((2\sqrt{3})^3 = 8 \times 3\sqrt{3} = 24\sqrt{3}\)

Step 4: Calculate \(3(1)(2\sqrt{3})\)

\(3(1)(2\sqrt{3}) = 3 \times 2\sqrt{3} = 6\sqrt{3}\)

Step 5: Combine the terms to find \(x^3 - y^3\)

\(x^3 - y^3 = 24\sqrt{3} + 6\sqrt{3}\)

\(x^3 - y^3 = (24+6)\sqrt{3}\)

\(x^3 - y^3 = 30\sqrt{3}\)

Thus, the value of \(x^3 - y^3\) is \(30\sqrt{3}\).

Revision Table: Key Concepts for Radical Expressions and Cubes

Additional Information: Working with Algebraic Identities

Algebraic identities are equations that are true for all possible values of the variables involved. They are very useful for simplifying expressions and solving equations.

• The difference of squares identity, \((a+b)(a-b) = a^2 - b^2\), was crucial in rationalizing the denominators of \(x\) and \(y\).
• The difference of cubes identity, \(a^3 - b^3 = (a-b)^3 + 3ab(a-b)\), allowed us to calculate the final value efficiently using the calculated values of \(x-y\) and \(xy\). Another common form is \(a^3 - b^3 = (a-b)(a^2+ab+y^2)\). You could also calculate \(x^2\) and \(y^2\) separately and use this form.
• Calculating \(x^2\): \((2+\sqrt{3})^2 = 2^2 + 2(2)(\sqrt{3}) + (\sqrt{3})^2 = 4 + 4\sqrt{3} + 3 = 7 + 4\sqrt{3}\)
• Calculating \(y^2\): \((2-\sqrt{3})^2 = 2^2 - 2(2)(\sqrt{3}) + (\sqrt{3})^2 = 4 - 4\sqrt{3} + 3 = 7 - 4\sqrt{3}\)
• Using the formula \(x^3 - y^3 = (x-y)(x^2+xy+y^2)\):
  Substitute \(x-y = 2\sqrt{3}\), \(xy=1\), \(x^2 = 7+4\sqrt{3}\), \(y^2 = 7-4\sqrt{3}\).
  \(x^3 - y^3 = (2\sqrt{3})((7+4\sqrt{3}) + 1 + (7-4\sqrt{3}))\)
  \(x^3 - y^3 = (2\sqrt{3})(7 + 4\sqrt{3} + 1 + 7 - 4\sqrt{3})\)
  \(x^3 - y^3 = (2\sqrt{3})((7+1+7) + (4\sqrt{3}-4\sqrt{3}))\)
  \(x^3 - y^3 = (2\sqrt{3})(15 + 0)\)
  \(x^3 - y^3 = (2\sqrt{3})(15) = 30\sqrt{3}\)
  Both methods give the same result, confirming the calculation.

Understanding how to manipulate expressions with radicals and apply algebraic identities is fundamental in solving such problems.