Question

(3a-1/a) ³ cube in expand form

Answer 1

Answer:

The answer is $27a^3-27a+\frac{9}{a}-\frac{1}{a^3}$

Step-by-step explanation:

Using $(a-b)^3=a^3-b^3-3ab(a-b)$

The expanded form of   $(3a-\frac{1}{a})^3$

                                       $=(3a)^3-(\frac{1}{a})^3-3\times 3a\times \frac{1}{a}(3a-\frac{1}{a})\\=27a^3-\frac{1}{a^3}-9(3a-\frac{1}{a})\\=27a^3-\frac{1}{a^3}-27a+\frac{9}{a}\\=27a^3-27a+\frac{9}{a}-\frac{1}{a^3}$