Finding even degree of a polynomial using even function trick

Question

The sum of the coefficents of all even degree terms in x in the expansion of $(x + \sqrt{x^3 - 1})^6 +( x - \sqrt{x^3 -1})^6 $ , ($x>1$) is equal to:

ans:24

I tried to do this by the trick to extract out even part of a function, but I am getting the wrong answer:

$$ f_{even} (1) = \frac{ f(1) + f(-1) }{2} = \frac{\left[2 \right] + \left[ (1-i\sqrt{2})^6 + (1+i \sqrt{2})^6 \right]}{2}= 1+ \frac{3^3}{2}\left[ \binom{6}{0} -2^2 \binom{6}{2} + 2^4 \binom{6}{4} - 2^6 \binom{6}{0} \right] \tag{1} $$

Now,

$$\binom{6}{0} -2^2 \binom{6}{2} + 2^4 \binom{6}{4} - 2^6 \binom{6}{0} = 1-15 \left[16-4 \right] - 64$$

The above is a very negative value no way will it give 24 when we add it one. So, why does extracting even part of a function trick fail here for finding the answer?

To be explicit, I'm talking about this identity

$$ f_{even} (x)= \frac{f(x) + f(-x)}{2}$$

Note: I evaluated $f(-1)$ in the following way:

$$ f(-1) = (-1+\sqrt{-1^3-1})^6 + (-1 - \sqrt{-1^3 -1 })^6= (-1 + i \sqrt{2})^6 + (1 + i \sqrt{2})^6 = (1-i \sqrt{2})^6 + (1+ i \sqrt{2})^6$$

Answer 1

The middle of your discussion should read:

$$ f_{even} (1) \ = \ \frac{ f(1) + f(-1) }{2} \ = \ \frac{\left[2 \right] + \left[ (1-i\sqrt{2})^6 + (1+i \sqrt{2})^6 \right]}{2}$$ $$ = \ 1 \ + \ \frac{1}{2} \ \left[ \ 2 · \binom{6}{0} \ + \ 2 · \binom{6}{2} · (i \sqrt2)^2 \ + \ 2 · \binom{6}{4} · (i \sqrt2)^4 \ + \ 2 · \binom{6}{6} · (i \sqrt2)^6 \ \right] $$

$$ = \ 1 \ + \ \left[ \ 1 \ + \ 15 · (-2) \ + \ 15 · (+4) \ + \ 1 · (-8) \ \right] \ \ = \ \ 1 + 1 - 30 + 60 -8 \ \ = \ \ 24 \ \ . $$

There doesn't seem to be anything wrong with your method (which agrees with the result from directly computing the polynomial $ \ 2 x^9 + 30 x^8 + 30 x^7 - 4 x^6 - 60 x^5 - 30 x^4 + 6 x^3 + 30 x^2 - 2 \ \ , $ which arises from your original expression). However, you are raising $ \ 2 \ $ to even powers, rather than $ \ \mathbf{\sqrt2} \ $ , and I don't understand where your factor of $ \ 3^3 \ $ comes from (if it's the modulus of $ \ 1 \pm \sqrt2 \ $ raised to the sixth power, it isn't needed in this calculation).

Answer 2

$$=2\sum_{r=0}^{2r\le6}\binom6{2r}x^{6-2r}(x^3-1)^r=2\left(x^6+\binom62x^4(x^3-1)+\binom64x^2(x^3-1)^2+(x^3-1)^3\right)$$

$(x^3-1)^2=\underbrace{1}\cdot x^6-2x^3+\underbrace{1}$

$(x^3-1)^3=x^9+\underbrace{(-3)}x^6+3x^3+\underbrace{(-1)}$

So, the sum of even degree terms in $x,$

$$2+2\binom62(-1)+2\binom64(1+1)+2(-3-1)$$