If the expansion of (x2 + ax + 8) (x2-3x + b) does not contain constant term and X3 term, the values of a and B are obtained

If the expansion of (x2 + ax + 8) (x2-3x + b) does not contain constant term and X3 term, the values of a and B are obtained

Original formula = x ^ 4 + (A-3) x & # 179; + (8-3a + b) x + (AB-24) x + 8B
If not, then A-3 = 0
8b=0
So a = 3, B = 0

If (x2 + ax + b) (x2 + 3x + 1) does not contain x2 and X terms, the values of a and B are obtained
Thank you Yiye Zhiqiu for your answer. I've been taught, but there is still a question. If you say so, the solution of a and B should be a = 3 / 8; b = 1 / 8
It's the same answer as the test paper,
From the Title Meaning: 3ax2 + bx2 + x2 = 0
Then 3A + B + 1 = 0
a+3b=0
The solution is a = 3 / 8;
B=1/8

(x2 + ax + b) (x2 + 3x + 1) = X4 + 3x3 + x2 + AX3 + 3ax2 + ax + bx2 + 3bx + B. If 3ax2 + bx2 + x2 = 0, then 3A + B = 1, ax + 3bx = 0, a = 3B 9b + B = 10B = 1, so B = 0.1, a = 3B = 0.3, without x2 and X terms

In the expansion of binomial (x ^ 2-1 / x) ^ n, if the sum of all binomial coefficients is 32, then the sum of all coefficients in the expansion is 32

In the expansion of binomial (x ^ 2-1 / x) ^ n, the sum of all binomial coefficients is 2 ^ n
SO 2 ^ n = 32 = 2 ^ 5
So n = 5
When x = 1, n = 5, (x ^ 2-1 / x) ^ n = 0
The sum of the coefficients in the expansion is 0