Question

6 Phân tích đa thức sau thành nhân tử:

a) x^3+8/27 ; b) 27x^3-1/8 ; c) (x+2y)^3-(x-2y)^3

Answer 1

\(a,x^3+\dfrac{8}{27}=x^3+\left(\dfrac{2}{3}\right)^3=\left(x+\dfrac{2}{3}\right)\left(x^2-\dfrac{4}{3}x+\dfrac{4}{9}\right)\)

\(b,27x^3-\dfrac{1}{8}=\left(3x\right)^3-\left(\dfrac{1}{2}\right)^3=\left(3x-\dfrac{1}{2}\right)\left(9x^2+3x+\dfrac{1}{4}\right)\)

\(c,\left(x+2y\right)^3-\left(x-2y\right)^3\)

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

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

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