Phân tích đa thức sau thành nhân tử: a3 + b3 + c3 – 3abc.

Trả lời

$\begin{array}{l}{a}^3+b^3+c^3–3abc={\left(a+b\right)}^3–3a^{3}b–3a{b}^3+c–3abc\\ =\left[{\left(a+b\right)}^3+c^3\right]-3ab\left(a+b+c\right)\\ =\left(a+b+c\right)\left[{\left(a+b\right)}^2-c\left(a+b\right)+c^2\right]-3ab\left(a+b+c\right)\\ =\left(a+b+c\right)(a^2+2ab+b^2-ca-bc+c^2-3ab)\\ =\left(a+b+c\right)(a^2+b^2+c^2-ab-bc-ca).\end{array}$