1) \(a\left(b^3-c^3\right)+b\left(c^3-a^3\right)+c\left(a^3-b^3\right)\)
\(=a\left(b^3-c^3\right)-b\left[\left(b^3-c^3\right)+\left(a^3-b^3\right)\right]+c\left(a^3-b^3\right)\)
\(\left(do\left[\left(b^3-c^3\right)+\left(a^3-b^3\right)\right]=-\left(c^3-a^3\right)\right)\)
\(=\left(a-b\right)\left(b^3-c^3\right)+\left(c-b\right)\left(a^3-b^3\right)\)
\(=\left(a-b\right)\left(b-c\right)\left(b^2+bc+c^2\right)-\left(b-c\right)\left(a-b\right)\left(a^2+ab+b^2\right)\)
\(=\left(a-b\right)\left(b-c\right)\left[\left(b^2+bc+c^2\right)-\left(a^2+ab+b^2\right)\right]\)
\(=\left(a-b\right)\left(b-c\right)\left[\left(c^2-a^2\right)+\left(bc-ab\right)\right]\)
\(=\left(a-b\right)\left(b-c\right)\left[\left(c-a\right)\left(c+a\right)+b\left(c-a\right)\right]\)
\(=\left(a-b\right)\left(b-c\right)\left(c-a\right)\left(a+b+c\right)\)
2) \(\dfrac{a-b}{b+c}+\dfrac{b-a}{c+a}+\dfrac{c-b}{a+b}=1\)
\(\Rightarrow\dfrac{a-c}{b+c}+1+\dfrac{b-a}{c+a}+1+\dfrac{c-b}{a+b}+1=4\)
\(\Rightarrow\dfrac{a-c+b+c}{b+c}+\dfrac{b-a+c+a}{c+a}+\dfrac{c-b+a+b}{a+b}=4\)
\(\Rightarrow\dfrac{a+b}{b+c}+\dfrac{b+c}{c+a}+\dfrac{c+a}{a+b}=4\)
