Giải:
1, $a^3+b^3+c^3=3abc \iff a+b+c=0$ v $a=b=c$
ĐK: $a \not = b \not = c \not =0$
$\iff a=-b-c \ \ ; b=-a-c \ ; \ c=-a-b$
Ta có: $\dfrac{a}{b-c}(\dfrac{b-c}{a}+\dfrac{c-a}{b}+\dfrac{a-b}{c})=1+\dfrac{a}{b-c}(\dfrac{c-a}{b}+\dfrac{a-b}{c})$
$= 1+\dfrac{a(c^2-ca+ab-b^2)}{(b-c)bc}=1+\dfrac{a(b-c)(a-b-c)}{(b-c)bc}=1+\dfrac{a(a-b-c)}{bc}=1+\dfrac{2a^2}{bc} \ (1)$
TT: $\dfrac{b}{c-a}(\dfrac{b-c}{a}+\dfrac{c-a}{b}+\dfrac{a-b}{c})=1+\dfrac{2b^2}{ac} \ (2)$ ; $\dfrac{c}{a-b}(\dfrac{b-c}{a}+\dfrac{c-a}{b}+\dfrac{a-b}{c})=1+\dfrac{2c^2}{ab} \ (3)$
(1)+(2)+(3) BTBĐ $=3+\dfrac{2a^2}{bc}+\dfrac{2b^2}{ca}+\dfrac{2c^2}{ab}=3+\dfrac{2a^3+2b^3+2c^3}{abc}=3+\dfrac{6abc}{abc}=9$ (đpcm)