Question

2a^-3/(2a)^-3

Answer 1

first you need to Arrange no. with alphabet and then go thru on latter and you can
solve it

Answer 2

i mean first you need to Arrange no. with alphabet and then go thru on Number and you can
solve it

Answer 3

1

Answer 4

\[\large \frac{2a^{-3}}{(2a)^{-3}}\] first expand \(\large (2a)^{-3}\)\[\large (2a)^{-3} = 2^{-3}\cdot a^{-3}\] so now evaluate what you can. \[\large \frac{2 \cdot \cancel{a^{-3}}}{2^{-3}\cdot \cancel {a^{-3}}}\]You're left with \[\large \frac{2}{2^{-3}}= 2^{1-(-3)} = 2^{4} =16\]

Answer 5

Another approach:
\[\frac{2a^{-3}}{(2a)^{-3}} = \frac{2(2a)^3}{a^3} = \frac{2*2^3*\cancel{a^3}}{\cancel{{a^3}}}=2*8=16\]because \[a^{-n}=\frac{1}{a^n}\]