Question

[Nữ hoàng sến súa là ta]

Phân tích đa thức sau thành nhân tử

\(h,\left(x-y\right)^3+\left(y-z\right)^3+\left(z-x\right)^3\)

\(i,x^4+4x^2+16\)

Answer 1

[Pham Van Hung]

  \(A=\left(x-y\right)^3+\left(y-z\right)^3+\left(z-x\right)^3\)

Đặt \(x-y=a,y-z=b,z-x=c\Rightarrow a+b+c=0\)

\(\Rightarrow a+b=-c\)

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

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

\(\Rightarrow a^3+b^3+3ab.\left(-c\right)=-c^3\)

\(\Rightarrow a^3+b^3+c^3=3abc\)

Vậy \(A=3\left(x-y\right)\left(y-z\right)\left(z-x\right)\)

        \(x^4+4x^2+16\)

\(=\left(x^2\right)^2+2.x^2.4+4^2-4x^2\)

\(=\left(x^2+4\right)^2-\left(2x\right)^2\)

\(=\left(x^2-2x+4\right)\left(x^2+2x+4\right)\)