cho a;b;c khác 0 và 1/a+1/b+1/c=0.Chứng minh rằng 1/a^3+1/b^3+1/c^3=3/abc

$\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=0$ 

$⇒\dfrac{1}{a}+\dfrac{1}{b}=-\dfrac{1}{c}$ 

$⇒(\dfrac{1}{a}+\dfrac{1}{b})^3=-\dfrac{1}{c^3}$ 

$⇒\dfrac{1}{a^3}+\dfrac{1}{b^3}+\dfrac{3}{ab}(\dfrac{1}{a}+\dfrac{1}{b})=-\dfrac{1}{c^3}$

$⇒\dfrac{1}{a^3}+\dfrac{1}{b^3}+\dfrac{3}{ab}.(-\dfrac{1}{c})=-\dfrac{1}{c^3}$ 

$⇒\dfrac{1}{a^3}+\dfrac{1}{b^3}-\dfrac{3}{abc}=-\dfrac{1}{c^3}$ 

$⇒\dfrac{1}{a^3}+\dfrac{1}{b^3}+\dfrac{1}{c^3}=\dfrac{3}{abc}$ 

$→$ đpcm

Đặt $x=\dfrac{1}{a},y=\dfrac{1}{b},z=\dfrac{1}{c}$

$=>x+y+z=0$

$=>x+y=-z$

$=>(x+y)^3+z^3=0$

$=>x^3+y^3+z^3=3xyz$

$=>\dfrac{1}{a^3}+\dfrac{1}{b^3}+\dfrac{1}{c^3}=\dfrac{3}{abc}$