Question

Find the cube of : $(2a+\frac{1}{2a});a\ne 0$

• A. $8a^3+6a+\frac{3}{2a}+\frac{1}{8a^3}$
• B. $8a^3+3a+\frac{3}{a}+\frac{1}{8a^3}$
• C. $8a^3+3a+\frac{6}{a}+\frac{1}{8a^3}$
• D. $8a^3+6a+\frac{6}{a}+\frac{1}{8a^3}$

Answer 1

The correct option is A $8a^3+6a+\frac{3}{2a}+\frac{1}{8a^3}$

Given equation is: $2a+\frac{1}{2a}$
Taking cube on both sides, we get
$(2a+\frac{1}{2a}{)}^3=(2a{)}^3+3\left(2a{)}^2\right(\frac{1}{2a})+3(2a\left)\right(\frac{1}{2a}{)}^2+(\frac{1}{2a}{)}^3$
$=8a^3+6a+\frac{3}{2a}+\frac{1}{8a^3}$