If α, β ≠ 0, and f(n) = αn + βn and | | | | |31+f(1)1+f(2)1+f(1)1+f(2)1+f(3)1+f(2)1+f(3)1+f(4)| | | | | = K(1−α)²(1−β)²(α−β)², then K is equal to

Let the given expression be

| | | | |31+f(1)1+f(2)1+f(1)1+f(2)1+f(3)1+f(2)1+f(3)1+f(4)| | | | |

where f(n) = αn + βn

The expression can be written as

| | | | |31 + (α + β)1 + (2α + 2β)1 + (α + β)1 + (2α + 2β)1 + (3α + 3β)1 + (2α + 2β)1 + (3α + 3β)1 + (4α + 4β)| | | | |

= | | | | |3 + α + β + 2α + 2β + α + β + 2α + 2β + 3α + 3β + 2α + 2β + 3α + 3β + 4α + 4β| | | | |

= | | | | |3 + 13α + 13β| | | | |

= |3 + 13(α + β)|

Given that

| | | | |31+f(1)1+f(2)1+f(1)1+f(2)1+f(3)1+f(2)1+f(3)1+f(4)| | | | | = K(1−α)²(1−β)²(α−β)²

Therefore,

|3 + 13(α + β)| = K(1−α)²(1−β)²(α−β)²

However, the given solution is incomplete and does not provide a way to determine the value of K. The determinant expression needs to be expanded and simplified to relate it to the given equation. Without further information or a corrected determinant expression, the value of K cannot be determined.