Question

Phân tích đa thức thành nhân tử :

a) x⁸ - y⁸

b) x⁶ - 1

Answer 1

a, \(x^8-y^8=\left(x^4\right)^2-\left(y^4\right)^2=\left(x^4-y^4\right)\left(x^4+y^4\right)\)

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

b, \(x^6-1=\left(x^3-1\right)\left(x^3+1\right)\)

\(=\left(x-1\right)\left(x^2+x+1\right)\left(x+1\right)\left(x^2-x+1\right)\)

Answer 2

a) \(x^8-y^8\Leftrightarrow\left(x^4\right)^2-\left(y^4\right)^2\Leftrightarrow\left(x^4+y^4\right)\left(x^4-y^4\right)\)

\(\Leftrightarrow\left(x^4+y^4\right)\left(\left(x^2\right)^2-\left(y^2\right)^2\right)\Leftrightarrow\left(x^4+y^4\right)\left(x^2+y^2\right)\left(x^2-y^2\right)\)

\(\Leftrightarrow\left(x^4+y^4\right)\left(x^2+y^2\right)\left(x+y\right)\left(x-y\right)\)

b) \(x^6-1\Leftrightarrow\left(x^2\right)^3-1\Leftrightarrow\left(x^2-1\right)\left(\left(x^2\right)^2+x^2.1+1^2\right)\)

\(\Leftrightarrow\left(x+1\right)\left(x-1\right)\left(x^4+x^2+1\right)\Leftrightarrow\left(x+1\right)\left(x-1\right)\left(x^4+2x^2+1-x^2\right)\)

\(\Leftrightarrow\left(x+1\right)\left(x-1\right)\left(\left(x^2+1\right)^2-x^2\right)\Leftrightarrow\left(x+1\right)\left(x-1\right)\left(x^2+1+x\right)\left(x^2+1-x\right)\)