Giải thích các bước giải:
Ta có:
$y^3-1=(y-1)(y^2+y+1)$
Vì $x+y=1\to y-1=-x$
$\to y^3-1=-x(y^2+y+1)$
$\to \dfrac{x}{y^3-1}=\dfrac{x}{-x(y^2+y+1)}=\dfrac{-1}{y^2+y+1}$
Tương tự $\dfrac{y}{x^3-1}=\dfrac{-1}{x^2+x+1}$
$\to \dfrac{x}{y^3-1}-\dfrac{y}{x^3-1}=\dfrac{-1}{y^2+y+1}-\dfrac{-1}{x^2+x+1}$
$\to \dfrac{x}{y^3-1}-\dfrac{y}{x^3-1}=\dfrac{-1}{y^2+y+1}+\dfrac{1}{x^2+x+1}$
$\to \dfrac{x}{y^3-1}-\dfrac{y}{x^3-1}=\dfrac{1}{x^2+x+1}-\dfrac{1}{y^2+y+1}$
$\to \dfrac{x}{y^3-1}-\dfrac{y}{x^3-1}=\dfrac{y^2+y+1-(x^2+x+1)}{(x^2+x+1)(y^2+y+1)}$
$\to \dfrac{x}{y^3-1}-\dfrac{y}{x^3-1}=\dfrac{(y^2-x^2)+(y-x)}{(x^2+x+1)(y^2+y+1)}$
$\to \dfrac{x}{y^3-1}-\dfrac{y}{x^3-1}=\dfrac{(y-x)(y+x)+(y-x)}{(x^2+x+1)(y^2+y+1)}$
$\to \dfrac{x}{y^3-1}-\dfrac{y}{x^3-1}=\dfrac{(y-x)\cdot 1+(y-x)}{(x^2+x+1)(y^2+y+1)}$
$\to \dfrac{x}{y^3-1}-\dfrac{y}{x^3-1}=\dfrac{2(y-x)}{(x^2+x+1)(y^2+y+1)}$
Ta có:
$(x^2+x+1)(y^2+y+1)$
$=x^2y^2+xy(x+y)+x^2+y^2+xy+(x+y)+1$
$=x^2y^2+xy\cdot 1+x^2+y^2+xy+1+1$
$=x^2y^2+xy+x^2+y^2+xy+2$
$=x^2y^2+x^2+y^2+2xy+2$
$=x^2y^2+(x+y)^2+2$
$=x^2y^2+1+2$
$=x^2y^2+3$
$\to \dfrac{x}{y^3-1}-\dfrac{y}{x^3-1}=\dfrac{2(y-x)}{x^2y^2+3}$
$\to \dfrac{x}{y^3-1}-\dfrac{y}{x^3-1}=-\dfrac{2(x-y)}{x^2y^2+3}$
$\to \dfrac{x}{y^3-1}-\dfrac{y}{x^3-1}+\dfrac{2(x-y)}{x^2y^2+3}=0$
$\to đpcm$
