Question:
Find the value
$(a-3 b)^{3}+(3 b-c)^{3}+(c-a)^{3}$
Solution:
Let $a-3 b=x, 3 b-c=y, c-a=z$
$x+y+z=a-3 b+3 b-c+c-a=0$
$(\therefore x+y+z=0) ?$
$x^{3}+y^{3}+z^{3}=3 x y z$
$=3(a-3 b)(3 b-c)(c-a)$
$\therefore(a-3 b)^{3}+(3 b-c)^{3}+(c-a)^{3}$
$=3(a-3 b)(3 b-c)(c-a)$
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