Tính:

$a) {(2x^2+3y)}^3;$

$b) {\left(\frac{1}{2}x-3\right)}^3;$

Giải:

$$\begin{gathered} a) {(2x^2+3y)}^3 \\ ={\left(2x^2\right)}^3+3{(2x^2}^2.3y+3.2x^2.{\left(3y\right)}^2+{\left(3y\right)}^3 \\ =8x^6+3.4x^4.3y+3.2x^2.9y^3+27y^3 \\ =8x^6+36x^{4}y+54x^2{y}^3+27y^3 \end{gathered}$$

$$\begin{gathered} b) {\left(\frac{1}{2}x-3\right)}^3 \\ ={\left(\frac{1}{2}x\right)}^3-3{\left(\frac{1}{2}x\right)}^2.3+3\left(\frac{1}{2}x\right).3^2-3^3 \\ =\frac{1}{8}{x}^3-\frac{9}{4}{x}^2+\frac{27}{2}x-27 \end{gathered}$$