The roots of the cubic equation 2x3 – 8x2 + ax + 6 = 0 are m + n, 2m, m – n. The value of a is ______.

To find the value of 'a', we need to use Vieta's formulas which state that the sum of the roots of a cubic equation ax³ + bx² + cx + d = 0 is -b/a, the sum of the product of the roots taken two at a time is c/a, and the product of the roots is -d/a.

Given the roots of the equation 2x³ - 8x² + ax + 6 = 0 are m + n, 2m, m - n.

1. The sum of the roots is (m + n) + 2m + (m - n) = 4m. According to Vieta's formulas, this should be equal to -b/a = 8/2 = 4. So, we have 4m = 4, which gives m = 1.

2. Substituting m = 1 into the roots, we get the roots as 1 + n, 2, 1 - n.

3. The sum of the product of the roots taken two at a time is (1 + n)2 + (1 + n)(1 - n) + 2*(1 - n). Simplifying this gives 2 + 2n + 1 - n² + 2 - 2n = 5 - n². According to Vieta's formulas, this should be equal to c/a. But we don't know the value of 'a' or 'c' yet, so we can't find the value of 'n' at this point.

4. The product of the roots is (1 + n)2(1 - n) = 2 - 2n². According to Vieta's formulas, this should be equal to -d/a = -6/2 = -3. So, we have 2 - 2n² = -3, which gives n² = 5/2 and n = sqrt(5/2) or n = -sqrt(5/2).

5. Substituting n = sqrt(5/2) into the equation from step 3, we get 5 - (5/2) = a. Simplifying this gives a = 5/2.

So, the value of 'a' is 5/2.