Given two equations $(xax)^3 = bx$ and $x^2a = (xa)^{-1}$ in a nonabelian group, solve for $x$. This is for a basic non-commutative group.
My steps:
$(xax)^3 = bx$
$xaxxaxxax = bx$
$xax^2ax^2ax = bx$
$xax^2ax^2a = b$
$x^2a = (xax^2a)^{-1}b$
Now, substituting $x^2a$ into the second equation:
$(xax^2a)^{-1}b = (xa)^{-1}$
$b = (xax^2a)(xa)^{-1} $
$b = (xax^2a)(a^{-1}x^{-1})$
$b = xax$
Now back to the first equation:
$b^3 = bx$
$b^2 = x$
Apparently, this is not the correct answer according to my answer sheet.
Am I making a mistake somewhere?
 A: The solution in the answer sheet is equivalent; your solution is fine.
Marking your work line by line . . .

My steps:
$(xax)^3 = bx$ $\color{red}{\quad\checkmark\text{Def}}$
$xaxxaxxax = bx$ $\color{red}{\quad\checkmark\text{Def}}$
$xax^2ax^2ax = bx$$\color{red}{\quad\checkmark\text{Rewrite}}$
$xax^2ax^2a = b$ $\color{red}{\quad\checkmark (\times x^{-1})}$
$x^2a = (xax^2a)^{-1}b$ $\color{red}{\quad\checkmark (\times (xax^2a)^{-1})}$
Now, substituting $x^2a$ into the second equation:
$(xax^2a)^{-1}b = (xa)^{-1}$ $\color{red}{\quad\checkmark (\text{Sub})}$
$b = (xax^2a)(xa)^{-1} $ $\color{red}{\quad\checkmark (\times (xax^2a))}$
$b = (xax^2a)(a^{-1}x^{-1})$ $\color{red}{\quad\checkmark (\text{Use of 'inverse of product' lemma})}$
$b = xax$ $\color{red}{\quad\checkmark (\text{Use of inverses})}$
Now back to the first equation:
$b^3 = bx$ $\color{red}{\quad\checkmark (\text{Sub})}$
$b^2 = x$ $\color{red}{\quad\checkmark (\times b^{-1})}$

A: You have $xaxxaxxax=bx$, so
$$
xa(x^2a)(x^2a)=b
$$
Now $a(x^2a)=a(xa)^{-1}=aa^{-1}x^{-1}=x^{-1}$ and therefore
$$
b=xx^{-1}a^{-1}x^{-1}=a^{-1}x^{-1}
$$
whence $x^{-1}=ab$ and $x=(ab)^{-1}$.
Can we say that $b^2=(ab)^{-1}$? This is equivalent to $b^3=a^{-1}$ or, as you proved that $b^3=bx$, to $bx=a^{-1}$, which is true.
A: *

*The following is  solution in the image attached from the link.
solution Image on paper
Form the above solution x = b-1a-1. substituting this in second equation given. a = b-3. which the solution b^2 can be written in infinite ways if you include a like b3ab2, b6a2b2 etc etc.
