To generate a Pythagorean triple using the polynomial identity \((x^2+y^2)^2 = (x^2 - y^2)^2 + (2xy)^2\), let's substitute \(x = 12\) and \(y = 9\):
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Calculate \(x^2\) and \(y^2\): \[ x^2 = 12^2 = 144 \] \[ y^2 = 9^2 = 81 \]
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Calculate \(x^2 + y^2\): \[ x^2 + y^2 = 144 + 81 = 225 \]
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Now, we calculate \((x^2 + y^2)^2\): \[ (x^2 + y^2)^2 = 225^2 = 50625 \]
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Next, calculate \(x^2 - y^2\): \[ x^2 - y^2 = 144 - 81 = 63 \] Then, calculate \((x^2 - y^2)^2\): \[ (x^2 - y^2)^2 = 63^2 = 3969 \]
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Now calculate \(2xy\): \[ 2xy = 2 \cdot 12 \cdot 9 = 216 \] Then, calculate \((2xy)^2\): \[ (2xy)^2 = 216^2 = 46656 \]
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Finally, check the original equation: \[ (x^2 + y^2)^2 = (x^2 - y^2)^2 + (2xy)^2 \] \[ 50625 = 3969 + 46656 \]
Now, since the original question only asks for one of the values of the Pythagorean triple, the two relevant parts we have derived from \(x = 12\) and \(y = 9\) are:
- One value is \(225\) (from \(x^2 + y^2\)).
- Another value is \(3969\) (from \((x^2 - y^2)^2\)).
- The calculated values also show \(216\) squared could be relevant.
Among the choices given in your question, the value 225 is indeed one of the values of the Pythagorean triple derived from \(x = 12\) and \(y = 9\).
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