Complex Multiplication!

We claim that at least one of $(a+b)$ , $(a+c)$ and $(a+d)$ is equal to zero. So, $(a+b)(a+c)(a+d)$ is always equal to zero.

Now we're going to prove this.

Notice that $a^3+b^3=-(c^3+d^3)$

$\Rightarrow (a+b)(a^2-ab+b^2)=-(c+d)(c^2-cd+d^2)$

As $-(c+d)=a+b$ , this equation can be rewritten as:

$(a+b)(a^2-ab+b^2)=(a+b)(c^2-cd+d^2)$

Now if $a+b=0$ , we're done. If it's not equal to zero, we can divide both sides of the equation by $(a+b)$ to get:

$a^2-ab+b^2=c^2-cd+d^2 \cdots (1)$

$\Rightarrow (a+b)^2-3ab=(c+d)^2-3cd$ .

Since $a+b=-(c+d)$ , $(a+b)^2=(c+d)^2$ and that means $ab=cd$ .

Go back to $(1)$ and re-write it as:

$a^2+b^2=c^2+d^2$

$\Rightarrow (a^2-c^2)+(b^2-d^2)=0$

$\Rightarrow (a+c)(a-c)+(b+d)(b-d)=0$

Substitute $b+d$ with $-(a+c)$ :

$(a+c)(a-c)-(a+c)(b-d)=0$

$\Rightarrow (a+c)(a-c-b+d)=0$

Since $-c-b=a+d$ , we can rewrite the whole thing as:

$2(a+c)(a+d)=0$

This means at least one of $(a+c)$ and $(a+d)$ is equal to zero.

So, $(a+b)(a+c)(a+d)=0$ and that is the only possible value of that.

That means the sum of all possible values of the given expression is equal to $0+0i$ .

And $p+q=\boxed{0}$ .

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I honestly have no idea what ITT is. I was reading a seminar lecture and came across a problem similar to this.

I will be using a lot of sigma summations in this solution, just a heads up.

Interestingly, $\sum_{cyc} a=\sum_{cyc} a^3=0$ leads to a lot of other zeros.

It turns out if you expand the desired product you get: $a^3+a^2(b+c+d)+\sum_{cyc} abc=\sum_{cyc} abc$ , which is a symmetric sum.

Now consider two constructed products:

$(\sum_{cyc} a)(\sum_{cyc} a^2)=\sum_{cyc} a^3+\sum_{cyc} a^2b$

$\Rightarrow \sum_{cyc} a^2b=0$

$(\sum_{cyc} a)^3=\sum_{cyc} a^3+3\sum_{cyc} a^2b+6\sum_{cyc} abc$

$\Rightarrow \sum_{cyc} abc=-\frac {\sum_{cyc} a^2b}{2}=0$ , and that's our answer.

According to me its bit obivous that two of things a b c d must be equal to 1 and other two equals to -1 thus... let us say that a =b=1 and c=d=-1 to get the equality that a+b+c+d=â^3+b^3+c^3+d^3=0 which satisfies my asumption so directly puting values.. (a+b)(a+c)(a+d)= (1+1)(1+-1)(1+-1) =0 thus p+qi=0+0i thus the answer is 0 ... Thats it