$a > b, (a - b)(b - c)(c - a) > 0$ Which is bigger, $a$ or $c$? $a > b$
$(a - b)(b - c)(c - a) > 0$
Which is bigger, $a$ or $c$?
 A: $a-b$ is positive, so $b-c$ and $c-a$ have the same sign. Then
$$a<c<b\text{ or }b<c<a$$
Since $b<a$ then the first option is false, so $a>c$.
A: Let us assume $c>a>b$. Then, we have $(a-b)(b-c)(c-a)<0$ because of $a-b>0,b-c<0,c-a>0$. 
Therefore we can rule out $c>a$. 
$a=c$ would imply $(a-b)(b-c)(c-a)=0$, so is impossible as well.
Therefore we can conclude $c<a$
A: Let $a\lt c$
It follows that $b\gt c$ so  that the initial condition
$(a-b)(b-c)(c-a)\gt 0$ holds true. ( If two of the terms within the parentheses are positive, the other one must be as well )
But $a\gt b$. A contradiction. Thus $a\gt c$.
A: Remember $k < j; x > 0$ then $kx < jx$ and $k/x < j/x$ and if $k < j; y < 0$ then $ky > jy$ and $k/y > j/y$.
Or in imformal terms "multiplying or dividing by a positive keeps the less than/greater than sign; multiplying or dividing by a negative flips it".
$a > b$ so $a - b>0$.
So if $(a-b)(b -c)(c-a) > 0$ then dividing be $a-b$ will "keep the sign"
$(b -c)(c-a) > 0$.  So either  $(b-c)$ and $(c-a)$ are both positive; or they are both negative.
If they are both positive then $b > c$ and $c > a$.  SO $b > c > a$ and $b > a$.  But we know that is not true.
If the are both negative then $b<c$ and $c < a$ and $a > b$ which is consistant with what we already know
So:  $a > b; c>a;$ and $c>b$ so
$(a-b) > 0$
$(a-b)(b-c) < 0$ and 
$(a-b)(b-c)(c-a) > 0$.
