Phân tích các đa thức sau thành nhân tử:

$$\begin{gathered} a. x^3+\frac{1}{27}; \\ b. {(a+b)}^3-{(a-b)}^3; \\ c. {(a+b)}^3+{(a-b)}^3; \\ d)8x^3+12x^{2}y+6x{y}^2+y^3; \\ e)-x^3+9x^2-27x+27; \end{gathered}$$

Giải:

$$\begin{gathered} a. x^3+\frac{1}{27} \\ =x^3+{\left(\frac{1}{3}\right)}^3 \\ =\left(x+\frac{1}{3}\right)\left(x^2-x.\frac{1}{3}+{\left(\frac{1}{3}\right)}^2\right) \\ =\left(x+\frac{1}{3}\right)\left(x^2-\frac{1}{3}x+\frac{1}{9}\right) \end{gathered}$$

$$\begin{gathered} b. {(a+b)}^3-{(a-b)}^3 \\ =\left[\right(a+b)-(a-b\left)\right][{(a+b)}^2+(a+b).(a-b)+{(a-b)}^2] \\ =(a+b-a+b)(a^2+2ab+b^2+a^2-b^2+a^2-2ab+b^2) \\ =2b(3a^3+b^2) \end{gathered}$$

$$\begin{gathered} c. {(a+b)}^3+{(a-b)}^3 \\ =\left[\right(a+b)+(a-b\left)\right][{(a+b)}^2-(a+b\left)\right(a-b)+{(a-b)}^2] \\ =(a+b+a-b)(a^2+2ab+b^2-a^2+b^2+a^2-2ab+b^2] \\ =2a.(a^2+3b^2) \end{gathered}$$

$$\begin{gathered} d)8x^3+12x^{2}y+6x{y}^2+y^3 \\ ={\left(2x\right)}^3+3.{\left(2x\right)}^2.y+3.2x.y+y^3 \\ ={(2x+y)}^3 \end{gathered}$$

$$\begin{gathered} e)-x^3+9x^2-27x+27 \\ =27-27x+9x^2-x^3 \\ =3^3-3.3^2.x+3.3.x^2-x^3 \\ ={(3-x)}^3 \end{gathered}$$