Question

[SSu_NNấm 241]

Phân tích thành nhân tử :
a, \(a\left(b^2+c^2\right)+b\left(a^2+c^2\right)+c\left(a^2+b^2\right)+2abc\)

b, \(a^3-b^3-c^3-3abc\)

Answer 1

[JOKER_Võ Văn Quốc]

a)a(b²+c²)+b(a²+c²)+c(a²+b²)+2abc

=ab²+ac²+ba²+bc²+ca²+cb²+2abc

=(ab²+ba²)+(ac²+bc²)+(ca²+abc)+(cb²+abc)

=ab(a+b)+c²(a+b)+ca(a+b)+cb(a+b)

=(a+b)(ab+c²+ca+cb)

=(a+b)(a+c)(b+c)

b)a³-b³-c³-3abc

=(a-b)³-c³+3ab(a-b)-3abc

=(a-b-c)[(a-b)²+(a-b)c+c²]+3ab(a-b-c)

=(a-b-c)(a^2-2ab+b²+ac-bc+c²+3ab)

=(a-b-c)(a²+b²+c²+ab-bc+ca)