To generate a Pythagorean triple using the polynomial identity \((x^2 + y^2)^2 = (x^2 - y^2)^2 + (2xy)^2\), we first need to calculate the components of the identity using \(x = 8\) and \(y = 6\).
-
Calculate \(x^2\) and \(y^2\): \[ x^2 = 8^2 = 64 \] \[ y^2 = 6^2 = 36 \]
-
Calculate \(x^2 + y^2\): \[ x^2 + y^2 = 64 + 36 = 100 \]
-
Calculate \((x^2 + y^2)^2\): \[ (x^2 + y^2)^2 = 100^2 = 10000 \]
-
Calculate \(x^2 - y^2\): \[ x^2 - y^2 = 64 - 36 = 28 \]
-
Calculate \(2xy\): \[ 2xy = 2 \cdot 8 \cdot 6 = 96 \]
Now we can form the Pythagorean triple using the results:
-
The first component of the triple: \[ (x^2 - y^2) = 28 \]
-
The second component: \[ (2xy) = 96 \]
-
The hypotenuse: \[ \sqrt{(x^2 + y^2)} = \sqrt{100} = 10 \]
Putting these together, we have the Pythagorean triple: \(28, 96, 100\).
To write the Pythagorean triple in ascending order: \[ (10, 28, 96) \]
However, \(10\) is not part of the original calculation of the triple, and this is not the right approach for forming a Pythagorean triple. The actual values from the identity must be used correctly.
Correctly formed, we actually achieve: \[ (28, 96, 100) \]
So, in ascending order: \[ (28, 96, 100) \]
Thus, the final answer is: \[ {28, 96, 100} \]
1 answer