Sum these coefficients

We can have the sum of the cofficients of f(x) without expanding out the expression for f(x) given in the problem. This because of the sum of the coefficients of a polynomial is the value of that polynomial for x=1. So we have f(1)=3^4 = 81

Sum of Coefficients = f(1) = 81

It's a fact that the sum of the coefficients of a given polynomial P(x) is P(1). Substituting 1 into the equation would give us,

$f(1) = (3[1^3] - 3[1] - 1)^{38} \times (2(1) + 1)^4 \times (1^4 - 3(1^2) + 5(1) - 2)^{77}$

$f(1) = ( [-1^{38}] \times [3^4] \times [1^{77}]$

$f(1) = \boxed {81}$

When we want to sum coefficients of polynomial $f(x)=a_nx^n+a_{n-1}x^{n-1}+...+a_1x+a_0$ , easiest way is to determine $f(1)$ , because $f(1)=a_n1^n+a_{n-1}1^{n-1}+...+1a_1+a_0=a_n+a_{n-1}+...+a_1+a_0$ , exactly what we want. Now to our task. We have our polynomial written as product of 3 polynomials. Result will be polynomial with degree of 38+4+77=119. But you don't have to multiply all those numbers. It would take you too much time. Just pluck $x=1$ into this "factored" form. Outcome will be the same. So, when $x=1$ we have: $(3\times 1^3-3\times 1-1)^{38} \times (2\times 1+1)^4 \times (1^4-3\times 1^2+5\times 1-2)^77 =$ $=(-1)^38 \times (3)^4 \times (1)^77=1\times 81\times 81=81$

Let x = 1 then ull get the answer

the sum of the coefficients of the first quantity is -1 which when raised to a positive even power gives positive 1. The sum of the coefficients of the second quantity is 3 which when raised to the 4th power gives 81. For the last quantity, the sum of the coefficients is 1 and when raised to any positive integer gives 1. The product of 1, 81 and 1 is 81.

Encontramos a soma dos coeficientes de um polinômio qualquer calculando $f(1)$ , segue,

$\\ f(1) = (3 \cdot 1^3 - 3 \cdot 1 - 1)^{38} \cdot (2 + 1)^4 \cdot (1 - 3 + 5 - 2)^{77} \\\\ f(1) = 1^{38} \cdot 3^4 \cdot 1^{77} \\\\ f(1) = 1 \cdot 81 \cdot 1 \\\\ \boxed{f(1) = 81}$

f(1)= the sum of the coefficients, e.g. let f(x) = x^2+4x+4 = (x+2)^2. therefore f(1) = (1)^2 + 4(1) + 4 = 9 = sum of coefficients. so f(1) of question above = (3(1)^3 - 3(1) - 1)^38 x (2(1) + 1)^4 x ((1)^4 - 3(1)^2 + 5(1) - 2)^77 = 81

We can get sum of all coefficients by putting x=1 thus simply calculating, we get f(1)= 81

It's quite a brute force solution but, by expanding this expression using Wolfram Alpha, we can get something quite big. (http://www.wolframalpha.com/input/?_=1393460775966&i=%283x^3+-+3x+-+1%29^38++%282x+%2b+1%29^4++%28x^4+-+3x^2+%2b+5x+-+2%29^77&fp=1&incTime=true)

Next, we can remove all x^... by using a regex. Python code: "k = re.split('x\^?[0-9]*', s)" where s is the expanded expression. And sum it up all: "sum(int(x) for x in k)" and we got 81.