Let's change the subject (2n-m) is regarded as a letter, and the algebraic formula (2n-m) ^ 2-1 - (2n-m) ^ 3 + 2 (2n-m) is arranged according to the descending power of the letter (2n-m) 2. We know the polynomial (a + b) x ^ 4 + (b-2) x ^ 3-2 (A-1) x ^ 2 + ax-3 about X, excluding x ^ 3 term and x ^ 2 term. When we calculate x = - 1, the values of these polynomials are It is known that the polynomial (m-2) x ^ 8-N ^ n + 1 + 3x + n about X is a quartic trinomial about X, and the value of M and N can be obtained 4. The binomial - 2 / 3x ^ m y ^ n + 1 is a quintic binomial about X and y, and M is a prime number and N is a non negative number. Find the value of m ^ n

Let's change the subject (2n-m) is regarded as a letter, and the algebraic formula (2n-m) ^ 2-1 - (2n-m) ^ 3 + 2 (2n-m) is arranged according to the descending power of the letter (2n-m) 2. We know the polynomial (a + b) x ^ 4 + (b-2) x ^ 3-2 (A-1) x ^ 2 + ax-3 about X, excluding x ^ 3 term and x ^ 2 term. When we calculate x = - 1, the values of these polynomials are It is known that the polynomial (m-2) x ^ 8-N ^ n + 1 + 3x + n about X is a quartic trinomial about X, and the value of M and N can be obtained 4. The binomial - 2 / 3x ^ m y ^ n + 1 is a quintic binomial about X and y, and M is a prime number and N is a non negative number. Find the value of m ^ n

1.-(2N-M)^3+(2N-M)^2+2(2N-M)-1
two point five
The previous process has been given to you
3. Because it is a quartic trinomial of X, m-2 = 0, so m = 2
4. Because it is about the quintic monomials of X and y
So m + n = 5
M is prime and N is nonnegative
So m = 2, n = 3 or M = 3, n = 2
optimum