al-jabr-waal-muq-abla

For the first question, i think it should be
(b-c)² , (c-a)² , (a-b)² are in AP...

Soln :

(b-c)² , (c-a)² , (a-b)² are in AP...
so,
(c-a)² - (b-c)² = (a-b)² - (c-a)²
or, (b-a)(2c-a-b) = (c-b)(2a-b-c) ...................................(i)

Now,
1b-c , 1c-a , 1a-b are in AP,
iff, 1c-a - 1b-c = 1a-b - 1c-a
iff, a+b-2c(c-a)(c-b) = b+c-2a(a-b)(c-a)

iff, (a-b)(a+b-2c) = (c-b)(b+c-2a)

This is true by (i).
Therefore, 1b-c , 1c-a , 1a-b are in AP

Another way of doing it is:

(b-c)^2, (c-a)^2, (a-b)^2 are in AP \Rightarrow \frac{(b-c)^2}{(a-b)(b-c)(c-a)}, \frac{(c-a)^2}{(a-b)(b-c)(c-a)}, \frac{(a-b)^2}{(a-b)(b-c)(c-a)} are in AP

\Rightarrow \frac{b-c}{(a-b)(c-a)}, \frac{c-a}{(a-b)(b-c)}, \frac{a-b}{(b-c)(c-a)} are in AP

\Rightarrow -\left[\frac{1}{a-b} + \frac{1}{c-a} \right], etc. are in AP

and for the final step, adding \left[\frac{1}{a-b} + \frac{1}{b-c} + \frac{1}{c-a} \right] to the three terms, we have

\frac{1}{b-c}, \frac{1}{c-a}, \frac{1}{a-b} are in AP

arey abhishek thats an easy problem....infact i just wanted to know how the "author" knew from beginning itself that among FOUR possible cases only these two will yield "all" the roots and roots of remaining cases will be covered up with these two cases only....or he just did not write it after solving...in short i wanted to know if there was any trick to know which cases are COVERING up all the roots?