Question

[Phạm Đỗ Bảo Ngọc]

phân tích đa thức sau thành nhân tử

a, x mũ 3 - 8

b, 27x mũ 3 + 125y mũ 3 

c, ( 2x - 1 ) mũ 3 + 8

d, x mũ 6 + 216

e, 1 - 27x mũ 3

j, ( x- 3 ) mũ 3 + 27

g, x mũ 3 y mũ 3 + 125

t, 8x mũ 3 - 1 phần 8

u, x mũ 3 + 1 phần 27

Answer 1

[Nguyễn Minh Quang]

a.\(x^3-8=x^3-2^3=\left(x-2\right)\left(x^2+2x+4\right)\)

b.\(27x^3+125y^3=\left(3x\right)^3+\left(5y\right)^3=\left(3x+5y\right)\left(9x^2-15xy+25y^2\right)\)

c.\(\left(2x-1\right)^3+8=\left(2x-1\right)^3+2^3=\left(2x+1\right)\left[\left(2x-1\right)^2-2\left(2x-1\right)+4\right]\)

d.\(x^6+6^3=\left(x^2+6\right)\left(x^4-6x+36\right)\)

e.\(1-27x^3=1-\left(3x\right)^3=\left(1-3x\right)\left(1+3x+9x^2\right)\)

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

g.\(x^3y^3+125=\left(xy\right)^3+5^3=\left(xy+5\right)\left(x^2y^2-5xy+25\right)\)

t.\(8x^3-\frac{1}{8}=\left(2x\right)^3-\left(\frac{1}{2}\right)^3=\left(2x-\frac{1}{2}\right)\left(4x^2+x+\frac{1}{4}\right)\)

u.\(x^3+\frac{1}{27}=x^3+\left(\frac{1}{3}\right)^3=\left(x+\frac{1}{3}\right)\left(x^2-\frac{x}{3}+\frac{1}{9}\right)\)