Question

Phân tích đa thức : a\(^3\)+b\(^3\)+c\(^3\)-3abc

Answer 1

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

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

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

Answer 2

a³ + b³ + c³ - 3abc

= \(\left(a^3+b^3\right)+c^3-3abc\)

= \(\left[\left(a+b\right)^3-3ab\left(a+b\right)\right]+c^3-3abc\)            (Đặt x = a + b)

= \(x^3-3abx+c^3-3abc\)

= \(x^3+c^3-3abx-3abc\)

= \(\left(x+c\right)\left(x^2+c^2-cx\right)-3ab\left(x+c\right)\)

= \(\left(x+c\right)\left(x^2+c^2-cx-3ab\right)\)                 (Thay x = a + b)

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

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

= \(\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ca\right)\)