Cách làm cho bạn:
a. $(2x^{2}+3y)^{3}=(2x^{2})^{3}+3(2x^{2})^{2}.3y+3.2x^{2}.9y^{2}+27y^{3}$
= $8x^{6} + 3 . 4x ^{4}. 3y + 3 . 2x ^{2}. 9y ^{2}+ 27y ^{3}$
= $8x ^{6}+ 36x^{4}y + 54x^{2}y^{2} + 27y^{3}$
Vậy $(2x^{2}+3y)^{3} = 8x ^{6}+ 36x^{4}y + 54x^{2}y^{2} + 27y^{3}$ .
b. $(\frac{1}{2}x-3)^{3}=(\frac{1}{2}x)^{3}-3(\frac{1}{2}x)^{2}.3+3\frac{1}{2}x.3^{2}-3^{3}$
= $\frac{1}{8}x^{3}-3\frac{1}{4}x^{2}.3+3\frac{1}{2}x.9-27$
= $\frac{1}{8}x^{3}-\frac{9}{4}x^{2}+\frac{27}{2}x-27$
Vậy $(\frac{1}{2}x-3)^{3} = \frac{1}{8}x^{3}-\frac{9}{4}x^{2}+\frac{27}{2}x-27$
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