If \(x^3 + \frac{1}{x^3} = \frac{65}{8}\) and \(y^3 + \frac{1}{y^3} = \frac{730}{27}\), then which one of the following is a value of \(xy\)?
Solving for \(xy\) Using Cubic Equations
This problem requires us to find the value of the product \(xy\) given two equations involving the sum of cubes: \(x^3 + \frac{1}{x^3} = \frac{65}{8}\) and \(y^3 + \frac{1}{y^3} = \frac{730}{27}\). We need to use algebraic identities to solve for \(x\) and \(y\) first.
Finding Possible Values for \(x\)
We know the algebraic identity: \(a^3 + b^3 = (a+b)^3 - 3ab(a+b)\). Applying this to \(x^3 + \frac{1}{x^3}\), we have \(a=x\) and \(b=\frac{1}{x}\).
So, \(x^3 + \frac{1}{x^3} = \left(x + \frac{1}{x}\right)^3 - 3(x)\left(\frac{1}{x}\right)\left(x + \frac{1}{x}\right)\).
This simplifies to \(x^3 + \frac{1}{x^3} = \left(x + \frac{1}{x}\right)^3 - 3\left(x + \frac{1}{x}\right)\).
Let's substitute \(u = x + \frac{1}{x}\). The equation becomes:
\( u^3 - 3u = \frac{65}{8} \)
To solve for \(u\), we can rearrange the equation:
\( 8u^3 - 24u = 65 \)
\( 8u^3 - 24u - 65 = 0 \)
We can look for rational roots. By testing values, we find that \(u = \frac{5}{2}\) is a solution:
\( 8\left(\frac{5}{2}\right)^3 - 24\left(\frac{5}{2}\right) - 65 = 8\left(\frac{125}{8}\right) - 12(5) - 65 = 125 - 60 - 65 = 125 - 125 = 0 \)
So, \(x + \frac{1}{x} = \frac{5}{2}\).
Now, we solve for \(x\):
\( x + \frac{1}{x} = \frac{5}{2} \)
Multiply by \(2x\) to clear denominators:
\( 2x^2 + 2 = 5x \)
Rearrange into a quadratic equation:
\( 2x^2 - 5x + 2 = 0 \)
Factor the quadratic:
\( (2x - 1)(x - 2) = 0 \)
This gives two possible values for \(x\): \(x = \frac{1}{2}\) or \(x = 2\).
Finding Possible Values for \(y\)
Similarly, we apply the same identity \(y^3 + \frac{1}{y^3} = \left(y + \frac{1}{y}\right)^3 - 3\left(y + \frac{1}{y}\right)\) to the second given equation.
Let \(v = y + \frac{1}{y}\). The equation becomes:
\( v^3 - 3v = \frac{730}{27} \)
Rearrange the equation:
\( 27v^3 - 81v = 730 \)
\( 27v^3 - 81v - 730 = 0 \)
By testing rational roots, we find that \(v = \frac{10}{3}\) is a solution:
\( 27\left(\frac{10}{3}\right)^3 - 81\left(\frac{10}{3}\right) - 730 = 27\left(\frac{1000}{27}\right) - 27(10) - 730 = 1000 - 270 - 730 = 1000 - 1000 = 0 \)
So, \(y + \frac{1}{y} = \frac{10}{3}\).
Now, we solve for \(y\):
\( y + \frac{1}{y} = \frac{10}{3} \)
Multiply by \(3y\) to clear denominators:
\( 3y^2 + 3 = 10y \)
Rearrange into a quadratic equation:
\( 3y^2 - 10y + 3 = 0 \)
Factor the quadratic:
\( (3y - 1)(y - 3) = 0 \)
This gives two possible values for \(y\): \(y = \frac{1}{3}\) or \(y = 3\).
Calculating the Value of \(xy\)
We have found the possible values for \(x\) are \(2\) and \(\frac{1}{2}\), and the possible values for \(y\) are \(3\) and \(\frac{1}{3}\). We need to consider all possible combinations for the product \(xy\):
- Case 1: If \(x = 2\) and \(y = 3\), then \(xy = 2 \times 3 = 6\).
- Case 2: If \(x = 2\) and \(y = \frac{1}{3}\), then \(xy = 2 \times \frac{1}{3} = \frac{2}{3}\).
- Case 3: If \(x = \frac{1}{2}\) and \(y = 3\), then \(xy = \frac{1}{2} \times 3 = \frac{3}{2}\).
- Case 4: If \(x = \frac{1}{2}\) and \(y = \frac{1}{3}\), then \(xy = \frac{1}{2} \times \frac{1}{3} = \frac{1}{6}\).
The possible values for \(xy\) are \(6\), \(\frac{2}{3}\), \(\frac{3}{2}\), and \(\frac{1}{6}\).
Conclusion
Comparing these possible values with the given options (3, 6, 8, 9), we see that \(6\) is one of the possible values for \(xy\).
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