<s>[INST] Let $a, b, c,x$ be elements of a group $G$. In each of the following, solve for $x$ in terms of $a, b, c$. Just checking my work. 

$axb =c.$
$(axb)a^{-1}b^{-1} = ca^{-1}b^{-1}$
$x = ca^{-1}b^{-1}$

$x^2b = xa^{-1}c$
$x^{-1}(x^2b) = (xa^{-1}c)x^{-1}$
$xb = a^{-1}c$
$x = a^{-1}b^{-1}c$

$x^2a = bxc^{-1}$ and $acx = xac.$
$(x^2a)x^{-1} = (bxc^{-1})x^{-1}$
$xa = bc^{-1}$
$(xa)c = (bc^{-1})c$
$acx = b$
$x = a^{-1}bc^{-1}$

$ax^2 = b$ and $x^3 = e.$
$(ax^2)x = bx$
$ae = bx$
$x = ab^{-1}$

$x^2 = a^2$ and $x^5 = e$
$(x^2)^2 = (a^2)^2$
$(x^4)x = a^4x$
$e = a^4x$
$(a^4)^{-1} = x$

$(xax)^3 = bx$ and $x^2a = (xa)^{-1}$
$((xax)^3)^{-2} = (bx)^{-2}$
$xax  = (bx)^{-2}$
$((xa)^{-1})^2 = ((bx)^{-2})^2$
$xa = 0$
$xaa^{-1} = 0a^{-1}$
$x = 0$
 [/INST] You’re forgetting that the group operation need not be commutative: the order of the operands matters. In the first problem, for example, $(axb)a^{-1}b^{-1}$ does not simplify to $x$; in fact, it doesn’t simplify at all without further knowledge of the group in question. You need to multiply $axb$ on the left by $a^{-1}$ and on the right by $b^{-1}$: 
$$a^{-1}(axb)b^{-1}=(a^{-1}a)x(bb^{-1})=x\;,$$
so $x=a^{-1}cb^{-1}$.
You’ve made similar errors in the second and third problems. The fourth is correct, as is the fifth, though its answer could better be expressed as $x=a^{-4}$. The last is wrong: $(y^3)^{-2}=y^{-6}$, not $y$.
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