What is the numerical value of the expression $\displaystyle \frac{(a + b)(b + c)(c + a)(8^{100}-7(8^{99})-7(8^{98})-\cdots-7(8))}{abc}$?

Let $a,\,b,\,c \in \Bbb{R}$ such that $\displaystyle \frac{a+b}{c}=\frac{b+c}{a}=\frac{c+a}{b}$

What is the numerical value of the expression $\displaystyle \frac{(a + b)(b + c)(c + a)(8^{100}-7(8^{99})-7(8^{98})-\cdots-7(8))}{abc}$?

My solution:

From the identity $a^n-1=(a-1)(a^{n-1}+a^{n-2}+\cdots+a+1)$, we can rewrite $\displaystyle 8^{100}-7(8^{99})-7(8^{98})-\cdots-7(8)$ as:

$\displaystyle \begin{align*}8^{100}-7(8^{99})-7(8^{98})-\cdots-7(8)&=8^{100}-7(8)(8^{98}+8^{97}+\cdots+1)\\&=8^{100}-\frac{7(8)(8-1)(8^{98}+8^{97}+\cdots+1)}{(8-1)}\\&=8^{100}-\frac{7(8)\color{yellow}\bbox[5px,purple]{(8-1)(8^{98}+8^{97}+\cdots+1)}}{7}\\&=8^{100}-8\color{yellow}\bbox[5px,purple]{(8^{99}-1)}\\&=8^{100}-8^{100}+8\\&=8\end{align*}$

Now, if we let $\displaystyle \frac{a+b}{c}=\frac{b+c}{a}=\frac{c+a}{b}=x$, we get

$\displaystyle a+b=cx,\,b+c=ca,\,c+a=bx$

Adding these above three gives:

$\displaystyle 2(a+b+c)=x(a+b+c)$

That means $x=2$ or $a+b+c=0, i.e. a+b=-c\implies \frac{a+b}{c}=-1=x$.

Therefore,

$\displaystyle \frac{(a + b)(b + c)(c + a)(8^{100}-7(8^{99})-7(8^{98})-\cdots-7(8))}{abc}$

$\displaystyle =\frac{(a + b)(b + c)(c + a)(8)}{abc}$

$\displaystyle =(x^3)(8)$

$\displaystyle =((-1)^3)(8)\stackrel{\text{or}}{=}((2)^3)(8)$

$\displaystyle =-8\stackrel{\text{or}}{=}64$