if 1/a+1/b+1/c=3 then prove that 1/a(b-1)(c-1) + 1/b(a-1)(c-1) + 1/c(a-1)(b-1) is 0.
Assume $\displaystyle a,b,c\not = 1$, otherwise the problem is ill defined.
Let $\displaystyle x=\frac{1}{a}, y=\frac{1}{b} , z=\frac{1}{c}$.
Then your original condition becomes $\displaystyle x+y+z=3$, and the expression you want to equate to zero becomes
$\displaystyle \frac{xyz}{(1-y)(1-z)}+\frac{xyz}{(1-x)(1-z)}+\frac{xyz}{(1-x)(1-y)}$
$\displaystyle =xyz\left(\frac{(1-x)+(1-y)+(1-z)}{(1-x)(1-y)(1-z)}\right)$
$\displaystyle =xyz\left(\frac{3-(x+y+z)}{(1-x)(1-y)(1-z)}\right)$
Substituting $\displaystyle x+y+z=3$ makes this whole expression zero.
You may have noticed I skipped a few steps here and there, but if you are doing these kind of questions you should be competent enough to work out the details.