Judging from your comment to Dilip's answer to your previous question you have not fully absorbed one aspect of finite field arithmetic. Namely the fact that if $\mathrm{char} K=p$ (or equivalently $q=p^m$ for some positive integer $m$) implies the rules $ (a+b)^p=a^p+b^p,\qquad (a-b)^p=a^p-b^p. $ I am willing to bet that this really is somewhere in your lecture notes, in which case this is just a refresher. Going from particular to general, let us first study the case $q=2^m$. Then $K$ has $\mathbb{F}_2=\mathbb{Z}/2\mathbb{Z}$ as a subfield, so in the field $K$ we have $2=1+1=0$, because that's the way it goes in that subfield. This implies that $ (a+b)^2=a^2+2ab+b^2=a^2+0ab+b^2=a^2+b^2 $ as claimed. Similarly, if $|K|=3^m$ it has $\mathbb{F}_3=\mathbb{Z}/3\mathbb{Z}$ as a subfield, and consequently $3=1+1+1=0$. This then implies $ (a+b)^3=a^3+3a^2b+3ab^2+b^3=a^3+0a^2b+0ab^2+b^3=a^3+b^3. $ The general case $|K|=p^m, p>3$ follows from a similar calculation with the binomial formula, after we first observe that the binomial coefficient $ {p\choose k}=\frac{p!}{(p-k)!k!} $ is divisible by $p$ (the numerator is manifestly divisible by $p$, but the denominator is not, because $p$ is a prime and all the factors in the factorials in the denominator are smaller than $p$). The claim for the differences then follows from this rule by replacing $b$ with $-b$.
As a consequence of this fact we also get that in any extension field of $K$, e.g. in $K(\alpha)$ we have $ (a-b)^q=(a-b)^{p^m}=\left((a-b)^p\right)^{p^{m-1}}=(a^p-b^p)^{p^{m-1}}=\cdots a^q-b^q $ proving the formula you had trouble with.