Question

Phân tích đa thức thành nhân tử :

a) (y³+8) +(y²-4)

b) x⁶-1

Answer 1

a)

\(\left(y^3+8\right)+\left(y^2-4\right)\)

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

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

\(=\left(y+2\right)\left(y^2+y+6\right)\)

b)

\(x^6-1=\)

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

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