Express roots of equation $acx^2-b(c+a)x+(c+a)^2=0$ in terms of $\alpha, \beta, $ 
If roots of the equation $ax^2+bx+c=0$ are $\alpha, \beta, $  find roots of equation $acx^2-b(c+a)x+(c+a)^2=0$ in terms of $\alpha, \beta$

Here's what I have tried so far,
I know that $\alpha+ \beta=\frac{-b}{a} $ and $\alpha \beta=\frac{c}{a} $
So I can express $b=-a(\alpha+\beta)$
$c=a.\alpha\beta$
Once I substitute for b and c in the equation I can get, $$\alpha\beta x^2+(\alpha+\beta)(\alpha\beta+1)x+(\alpha\beta+1)^2=0$$
I want to know whether there is any different approach other than this method?
Any hint is higly valued. thank you!

This is somewhat along the lines of the approach Quanto takes, but causes the roots to emerge in a different fashion.
Dividing the given quadratic equation $ \ ac·x^2 \ - \ b(c+a)·x \ + \ (c+a)^2 \ = \ 0 \ \ $ through by $ \ ac \ $ produces
$$ x^2 \ - \ \frac{b(c+a)}{ac}·x \ + \ \frac{(c+a)^2}{ac} \ = \ 0 \ \ \Rightarrow \ \ x^2 \ - \ \frac{b}{a} · \left(1+\frac{a}{c}\right)·x \ + \ \frac{c}{a} · \left(1+\frac{a}{c}\right)^2 \ = \ 0  \ \ . $$
Applying the Viete relations you cited, we can then express this as
$$ x^2 \ + \ (\alpha + \beta) · \left(1+\frac{1}{\alpha·\beta}\right)·x \ + \ (\alpha·\beta) · \left(1+\frac{1}{\alpha·\beta}\right)^2 \ = \ 0  \ \ . $$
Now it is a fairly familiar fact (which is simple enough to prove, though it is not often expressed this way) that if $ \ r \ $ and $ \ s \ $ are the roots of a quadratic equation $ \ x^2 + mx + n \ = \ 0 \ \ , $ then $ \ r + s \ = \ -m \ $ and $ \ r - s \ = \ \sqrt{\Delta} \ = \ \sqrt{m^2 - 4n} \ \ . $  For the equation at hand, we thus have
$$ r \ + \ s \ = \ - \left(\alpha \ + \ \beta \ + \ \frac{1}{\beta} \ + \ \frac{1}{\alpha} \right) \ \ , $$ $$ \Delta \ = \ (\alpha + \beta)^2 · \left(1+\frac{1}{\alpha·\beta}\right)^2 \ - \ \ 4  ·  (\alpha·\beta) · \left(1+\frac{1}{\alpha·\beta}\right)^2 \ \ = \ \ (\alpha - \beta)^2 · \left(1+\frac{1}{\alpha·\beta}\right)^2  $$
$$ \Rightarrow \ \ \sqrt{\Delta} \ \ = \ \ (\alpha - \beta) · \left(1+\frac{1}{\alpha·\beta}\right) \ \ = \ \ \left(\alpha \ - \ \beta \ + \ \frac{1}{\beta} \ - \ \frac{1}{\alpha} \right) \ \ . $$
Consequently,
$$ r \ \ = \ \ \frac12  ·  \left(-\alpha \ - \ \beta \ - \ \frac{1}{\beta} \ - \ \frac{1}{\alpha} \ + \ \alpha \ - \ \beta \ + \ \frac{1}{\beta} \ - \ \frac{1}{\alpha} \right) \ \ = \ \ - \left(\beta \ + \  \frac{1}{\alpha} \right) $$
and similarly,
$$ s \ \ = \ \ \frac12  ·  \left(-\alpha \ - \ \beta \ - \ \frac{1}{\beta} \ - \ \frac{1}{\alpha} \ - \ \alpha \ + \ \beta \ - \ \frac{1}{\beta} \ + \ \frac{1}{\alpha} \right) \ \ = \ \ - \left(\alpha \ + \  \frac{1}{\beta} \right) \ \ .  $$
