# What is the sum of the coefficients of ( [3x - 3x^2 +1]^744 ) x ( [- 3x + 3x^2 +1]^745 )?Though not particularly necessary, I'd personally appreciate it if you were to explain all of the steps and...

What is the sum of the coefficients of ( [3x - 3x^2 +1]^744 ) x ( [- 3x + 3x^2 +1]^745 )?

Though not particularly necessary, I'd personally appreciate it if you were to explain all of the steps and processes involved.

### 3 Answers | Add Yours

You should remember that you may evaluate the sum of coefficients of a polynomial if you substitute 1 for the variable x such that:

`(3x - 3x^2 +1)^744 = (3*1 - 3*1^2 + 1)^744`

`(3x - 3x^2 +1)^744 = (3 - 3 + 1)^744`

`(3*1 - 3*1^2 + 1)^744 = 1^744 = 1`

`(-3x + 3x^2 +1)^745 = (-3*1 + 3*1^2 + 1)^745`

`(-3x + 3x^2 +1)^745 = (-3 + 3 + 1)^745`

`(-3x + 3x^2 +1)^745 = 1^745 = 1`

`(3x - 3x^2 +1)^744*(-3x + 3x^2 +1)^745 = 1*1 = 1`

**Hence, evaluating the sum of coefficients and the given multiplication yields `(3x - 3x^2 +1)^744*(-3x + 3x^2 +1)^745 = 1.` **

F(x) = (1+3x-3x^2)^744 *(1-3x+3x^2)^745...(1)

Sum of the coefficient of x will be:

put x=1 in eq(1)

f(1) = (1+3-3)^744 * (1-3+3)^745=1

This is the sum of the coefficient that is = 1

a

**Sources:**

The given expression could be written as below:

f(x) = (1+3x-3x^2)^744 *(1-3x+3x^2)^745 = A0+A1x+Ax^2+A3x^3+...................................(1)

We are required to find the sum of the coefficients of x which is equal to A1+A2+A3+....

We put x = 1 in (1) and get:

f(1) = (1+3-3)^744 * (1-3+3)^745 = A0+A1+A2+A3+......

Therefore f(1) = 1^744 * 1^745 = **1 = A0+A1+A2+A3...**

**Therefore the sum of the coefficients of x in (1+3x-3x^2)^744 *(1-3x+3x^2)^745 is 1**.