To generate a Pythagorean triple using the polynomial identity \((x^2+y^2)^2 = (x^2-y^2)^2 + (2xy)^2\), we need to calculate the required terms using \(x = 18\) and \(y = 9\).
First, we calculate \(x^2\) and \(y^2\):
- \(x^2 = 18^2 = 324\)
- \(y^2 = 9^2 = 81\)
Now we can calculate the expressions:
- \(x^2 + y^2 = 324 + 81 = 405\)
- \((x^2 + y^2)^2 = 405^2 = 164025\)
Next, we calculate \(x^2 - y^2\) and \(2xy\):
- \(x^2 - y^2 = 324 - 81 = 243\)
- \(2xy = 2 \cdot 18 \cdot 9 = 324\)
Now we can find \((x^2 - y^2)^2 + (2xy)^2\):
- \((x^2 - y^2)^2 = 243^2 = 59049\)
- \((2xy)^2 = 324^2 = 104976\)
Adding these two results together:
\[ 59049 + 104976 = 164025 \]
Thus, both sides of the equation match, confirming that a Pythagorean triple can be generated.
The values of the generated Pythagorean triple are:
- The first side \(= x^2 + y^2 = 405\)
- The second side \(= |x^2 - y^2| = 243\)
- The hypotenuse \(= 2xy = 324\)
From the provided options (324, 162, 81, 729), the value that appears in our Pythagorean triple is:
324.
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