Remember that the cross product of a vector with itself is 0.

\(\displaystyle{i}\times{j}={k}\)

\(\displaystyle{j}\times{k}={i}\)

\(\displaystyle{k}\times{i}={j}\)

So, the given problem is \(\displaystyle{k}\times{\left({i}-{2}{j}\right)}={k}\times{i}-{2}{k}\times{j}\)

Replace \(\displaystyle-{2}{k}\times{j}\) with \(\displaystyle{2}{j}\times{k}\)

\(\displaystyle={k}\times{i}+{2}{j}\times{k}\)

\(\displaystyle={j}+{2}{i}\)

\(\displaystyle={2}{i}+{j}\)

Thus, we have \(\displaystyle{k}\times{\left({i}-{2}{j}\right)}={2}{i}+{j}\)

\(\displaystyle{i}\times{j}={k}\)

\(\displaystyle{j}\times{k}={i}\)

\(\displaystyle{k}\times{i}={j}\)

So, the given problem is \(\displaystyle{k}\times{\left({i}-{2}{j}\right)}={k}\times{i}-{2}{k}\times{j}\)

Replace \(\displaystyle-{2}{k}\times{j}\) with \(\displaystyle{2}{j}\times{k}\)

\(\displaystyle={k}\times{i}+{2}{j}\times{k}\)

\(\displaystyle={j}+{2}{i}\)

\(\displaystyle={2}{i}+{j}\)

Thus, we have \(\displaystyle{k}\times{\left({i}-{2}{j}\right)}={2}{i}+{j}\)