Find the relation between $a,b $ and $c$ in quadratic equation. 
If the roots of the equation $a(b-c)x^2+b(c-a)x+c(a-b)=0,\ \ \{a,b,c,x\}\in \mathbb{R}$ are equal, then $a,b,c$ are in

Options
$a.)\ AP\\
b.)\ GP\\
\color{green}{c.)\ HP}\\
d.)\ \text{cannot be determined}\\$
by using discriminant property
$[b(c-a)]^2-4ac(b-c)(a-b)=0$
I cannot reach any conclusion and is also cumbersome , though I don't know how wolfram
reached this
$[b(c-a)]^2-4ac(b-c)(a-b)=0\ \Longleftrightarrow \dfrac{2}{b}=\dfrac{1}{a}+\dfrac{1}{c}$
I look for a short and simple way .
I have studied maths up to $12$th grade.
 A: $$\begin{align}&b^2(c-a)^2=4ac(b-c)(a-b)\\&\Rightarrow b^2(c^2-2ac+a^2)=4ac(ab-b^2-ca+bc)\\&\Rightarrow b^2c^2-2ab^2c+a^2b^2=4a^2bc-4ab^2c-4a^2c^2+4abc^2\\&\Rightarrow b^2c^2+a^2b^2=4a^2bc-2ab^2c-4a^2c^2+4abc^2\end{align}$$Dividing the both sides by $a^2b^2c^2$, we have$$\begin{align}&\Rightarrow \frac{1}{a^2}+\frac{1}{c^2}=\frac{4}{bc}-\frac{2}{ac}-\frac{4}{b^2}+\frac{4}{ab}\\&\Rightarrow \frac{1}{a^2}-\frac{4}{ab}+\frac{4}{b^2}+\frac{1}{c^2}+\frac{2}{ac}-\frac{4}{bc}=0\\&\Rightarrow \left(\frac 1a-\frac 2b\right)^2+\frac 2c\left(\frac 1a-\frac 2b\right)+\frac{1}{c^2}=0\\&\Rightarrow \left(\frac 1a-\frac 2b+\frac 1c\right)^2=0\\&\Rightarrow \frac 1a-\frac 2b+\frac 1c=0\end{align}$$
A: solving this equation we get $$1=\frac{c(a-b)}{a(b-c)}$$ and this is equivalent to $$2ac=b(a+c)$$
A: Hint
Consider the function $$f(x)=a(b-c)x^2+b(c-a)x+c(a-b)$$ It is clear that $x=1$ is a solution ($f(1)=0$). 
Now, since you want the two roots of $f(x)=0$ to be equal, that is to say $x_1=x_2=1$, we then have $$x_1+x_2=2=-\frac{b(c-a)}{a(b-c)}$$ $$x_1 \times x_2=1=\frac{c(a-b)}{a(b-c)}$$ From  the first equation you can extract $b$ as a function of $a$ and $c$ to get $$b=\frac{2 a c}{a+c}$$ Plugging into the second equation, this gives the stupid $1=1$ result.
From here you arrive to the result $$\dfrac{2}{b}=\dfrac{1}{a}+\dfrac{1}{c}$$ that is to say $$b=\frac 2 {\frac 1a+\frac 1c}$$ so $b$ is the harmonic mean of $a$ and $c$.
