Question

Phân tích đa thức thành nhân tử:

a) 27x⁶ - 8x³

b) x⁶ - y⁶

c) y⁹ - 9x²y⁶ + 27x⁴y³ - 27x⁶

Answer 1

\(27x^6-8x^3=x^3\left(27x^3-8\right)=x^3\left[\left(3x\right)^3-2^3\right]=x^3\left(3x-2\right)\left(9x^2+6x+4\right)\)

\(x^6-y^6=\left(x^2\right)^3-\left(y^2\right)^3=\left(x^2-y^2\right)\left(x^4+x^2y^2+y^4\right)\)

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

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

\(y^9-9x^2y^6+27x^4y^3-27x^6=\left(y^3\right)^3-3.\left(y^3\right)^2.\left(3x^2\right)+3.y^3.\left(3x^2\right)^2-\left(3x^2\right)^3\)

\(=\left(y^3-3x^2\right)^3\)