Question

Phân tích đa thức x⁸ - 1 thành nhân tử

Answer 1

\(x^8-1=\left(x^4\right)^2-1^2=\left(x^4-1\right)\left(x^4+1\right)\)

\(=\left[\left(x^2\right)^2-1^2\right]\left(x^4+1\right)\)

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

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

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

Answer 2

x⁸-1⇔(x⁴)²-1⇔(x⁴+1)(x⁴-1)⇔(x⁴+1)(x²+1)(x²-1)

⇔(x⁴+1)(x²+1)(x+1)(x-1)