Verified answer

[tex]\displaystyle\bf\\1)\\\\\frac{1}{(a-b)(b-c)} -\frac{1}{(b-c)(a-c)} =\frac{a-c-a+b}{(a-b)(b-c)(a-c)} =\\\\\\=\frac{b-c}{(a-b)(b-c)(a-c)} =\frac{1}{(a-b)(a-c)} \\\\\\2)\\\\\frac{1}{(a-b)(a-c)} -\frac{1}{(c-a)(b-a)} =\frac{1}{(a-b)(a-c)} -\frac{1}{(a-c)(a-b)} =0\\\\\\0=0[/tex]

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