Question

rút gọn: (b-c)³+(c-a)³-(a-b)³-3(a-b)(b-c)(c-a)

Answer 1

Ta có: \(\left(b-c\right)^3+\left(c-a\right)^3-\left(a-b\right)^3-3\left(a-b\right)\left(b-c\right)\left(c-a\right)\)

\(=\left(b-c+c-a\right)\left[\left(b-c\right)^2-\left(b-c\right)\left(c-a\right)+\left(c-a\right)^2\right]-\left(a-b\right)\left[1+3\left(b-c\right)\left(c-a\right)\right]\)

\(=\left(b-a\right)\left(b^2-3bc+3c^2+ab-3ac+a^2\right)-\left(a-b\right)\left(1+3bc-3ab-3c^2+3ac\right)\)

\(=\left(b-a\right)\left(b^2-3bc+3c^2+ab-3ac+a^2+1+3bc-3ab-3c^2+3ac\right)\)

\(=\left(b-a\right)\left(b^2-2ab+a^2+1\right)\)

\(=\left(b-a\right)^3+\left(b-a\right)\)

\(=b^3-3b^2a+3ba^2-a^3+b-a\)