Game of Equations!

The answer is yes for part $A$ , no for part $B$ .

The first part is easy to prove. Note that when $c=-a-b-1$ , $x=1$ is a real solution to all three equations regardless of Brian's choice of $b$ .

To prove the second part, suppose that Brian pick $b$ such that $b=-a$ . Then, we have the following equations:

x^{3}+ax^{2}-ax+c=0 \ \ \ \tag{1}

x^{3}-ax^{2}+cx+a=0 \ \ \ \tag{2}

x^{3}+cx^{2}+ax-a=0 \ \ \ \tag{3}

$(1)+(2)=(2)+(3):$

$2x^{3}+(c-a)x+a+c=2x^{3}+(c-a)x^{2}+(a+c)x$

$(c-a)x^{2}+2ax-(a+c)=0$

$x=\frac{-2a \pm \sqrt{4a^{2}+4(c^{2}-a^{2})}}{2(c-a)}$

$x=\frac{-a \pm c}{c-a}$

Since $x <0, x \neq 1$ . Therefore, we change the $\pm$ sign to $-$ , implying $x=-\frac{a+c}{c-a}$ .

Since $(1)=(2)$ , we have

$x^{3}+ax^{2}-ax+c=x^{3}-ax^{2}+cx+a$

$2ax^{2}-(c+a)x+c-a=0$

Substituting $x=-\frac{a+c}{c-a}$ , we have

$\frac{2a(a+c)^{2}}{(c-a)^{2}}+\frac{(a+c)^{2}}{c-a}+c-a=0$

Multiply the equation throughout by $(c-a)^{2}$ , we get

$2a(a+c)^{2}+(a+c)^{2}(c-a)+(c-a)^{3}=0$

$(a+c)^{3}+(c-a)^{3}=0$

$a+c=-(c-a)$ , which implies $c=0$ .

Substituting this into $x=-\frac{a+c}{c-a}$ , we get $x=\frac{-a}{-a}=1$ . However, we have assumed that $x \neq 1$ earlier since we are looking for a negative $x$ which satisfies the equations!

The conclusion follows.