Let $a,\,b,\,c \in \Bbb{R}$ such that [MATH]\frac{a+b}{c}=\frac{b+c}{a}=\frac{c+a}{b}[/MATH]

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

My solution:

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

[MATH]\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*}[/MATH]

Now, if we let [MATH]\frac{a+b}{c}=\frac{b+c}{a}=\frac{c+a}{b}=x[/MATH], we get

[MATH]a+b=cx,\,b+c=ca,\,c+a=bx[/MATH]

Adding these above three gives:

[MATH]2(a+b+c)=x(a+b+c)[/MATH]

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

Therefore,

[MATH]\frac{(a + b)(b + c)(c + a)(8^{100}-7(8^{99})-7(8^{98})-\cdots-7(8))}{abc}[/MATH]

[MATH]=\frac{(a + b)(b + c)(c + a)(8)}{abc}[/MATH]

[MATH]=(x^3)(8)[/MATH]

[MATH]=((-1)^3)(8)\stackrel{\text{or}}{=}((2)^3)(8)[/MATH]

[MATH]=-8\stackrel{\text{or}}{=}64[/MATH]

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