Answer:
We can rewrite the given expressions as:
(3x² - y² - 2yz - z²) and (y² - z² - 2zx - x²)
To find the H.C.F., we can use the Euclidean algorithm. We start by dividing the first expression by the second expression:
(3x² - y² - 2yz - z²) ÷ (y² - z² - 2zx - x²)
Using long division or synthetic division, we get:
(3x² - y² - 2yz - z²) = (3x + y + z)(x - y + z) + 2y(x - z)
Therefore, the remainder is 2y(x - z). We can now divide the second expression by this remainder:
(y² - z² - 2zx - x²) ÷ 2y(x - z)
Using long division or synthetic division, we get:
(y² - z² - 2zx - x²) = -x(x - y + z) + z(x - y + z)
Therefore, the remainder is z(x - y + z).
Since the second remainder is not zero, we need to continue with the algorithm. Now we divide the remainder 2y(x - z) by the remainder z(x - y + z):
2y(x - z) ÷ z(x - y + z)
Using long division or synthetic division, we get:
2y(x - z) = 2y(x - y + z) - 2y²
Therefore, the remainder is -2y². Now we divide the previous remainder z(x - y + z) by this new remainder:
z(x - y + z) ÷ (-2y²)
Using long division or synthetic division, we get:
z(x - y + z) = -1(-2y²) + z²
Therefore, the H.C.F. of the original expressions is the absolute value of the last remainder, which is |-2y²| = 2y².
Therefore, the H.C.F. of (3x² - y² - 2yz - z²) and (y² - z² - 2zx - x²) is 2y².