Question

If $8a^{3b}$ = 1 and a > 0, find the value of $a^{2b} - \frac{1}{a^{b} }$

Answer 1

Some factoring lets us write

$a^{2b}-\dfrac1{a^b}=a^{2b}-a^{-b}=a^{-b}(a^{3b}-1)$

Then

$8a^{3b}=1\implies a^{3b}=\dfrac18$

$a^{2b}-\dfrac1{a^b}=a^{-b}\left(\dfrac18-1\right)=-\dfrac78a^{-b}=-\dfrac7{8a^b}$

Taking the cube root to solve for $b$, we find

$\sqrt[3]{a^{3b}}=\sqrt[3]{\dfrac18}\implies a^b=\dfrac12$

so ultimately

$a^{2b}-\dfrac1{a^b}=-\dfrac7{8\cdot\frac12}=-\dfrac74$

Answer 2

Answer:

The value of the given expression is    1 - 1 = 0

Step-by-step explanation:

8a^{3b} = 1   can be rewritten as (8a³)^b = 1 = (8a³)^0, which indicates that b = 0.  If b= 0, then 2b = 2(0) = 0.

Then a^(2b) = a^0 = 1, and 1/a^b = 1/1 = 1.

Thus, the value of the given expression is    1 - 1 = 0