Jane was tasked to simplify the expression $\left(\sqrt2+2\right)^6$ using the binomial theorem, which was written on the board.
Binomial Theorem: $$(a+b)^n=\sum_{k=0}^n{n\choose k}a^{n-k}\cdot b^k$$
(1) First, using the binomial theorem, Jane created the binomial expansion of the given expression. \begin{aligned} \left(\sqrt2+2\right)^6=\quad&{6\choose 0}\left(\sqrt2\right)^6\cdot2^0+{6\choose1}\left(\sqrt2\right)^5\cdot2^1+{6\choose2}\left(\sqrt2\right)^4\cdot2^2+\cr +&{6\choose3}\left(\sqrt2\right)^3\cdot2^3+{6\choose4}\left(\sqrt2\right)^2\cdot2^4+{6\choose5}\left(\sqrt2\right)^1\cdot2^5+{6\choose6}\left(\sqrt2\right)^0\cdot2^6 \end{aligned}
(2) In the next step, she evaluated the binomial coefficients and simplified the individual terms. \begin{aligned} &=\mathbf{0}\cdot\left(\sqrt2\right)^6+ \mathbf{6}\cdot\left(\sqrt2\right)^5\cdot2+\mathbf{15}\cdot\left(\sqrt2\right)^4\cdot4+\mathbf{20}\cdot\left(\sqrt2\right)^3\cdot8+\cr &\ \ \ \ \ +\mathbf{15}\cdot\left(\sqrt2\right)^2\cdot16+ \mathbf{6}\cdot\left(\sqrt2\right)^1\cdot32+\mathbf{1}\cdot\left(\sqrt2\right)^0\cdot64\cr &=12\cdot\left(\sqrt2\right)^5+60\cdot\left(\sqrt2\right)^4+160\cdot\left(\sqrt2\right)^3+240\cdot\left(\sqrt2\right)^2+192\cdot\sqrt2+64 \end{aligned}
(3) Then, Jane evaluated even exponents of $\sqrt2$ and summed the constants. \begin{aligned} &=12\cdot\left(\sqrt2\right)^5+\mathbf{60\cdot4}+160\cdot\left(\sqrt2\right)^3+\mathbf{240\cdot2}+192\cdot\sqrt2+64\cr &=12\cdot\left(\sqrt2\right)^5+160\cdot\left(\sqrt2\right)^3+192\cdot\sqrt2+784 \end{aligned}
(4) Next, Jane factored $\sqrt2$ out of the terms containing $\sqrt2$ with an odd exponent. $$=\sqrt2 \left(12\cdot\left(\sqrt2\right)^4+160\cdot\left(\sqrt2\right)^2+192\right)+784$$
(5) Again, she evaluated even exponents of $\sqrt2$ and simplified the expression to its final form: \begin{aligned} &=\sqrt2\cdot(12\cdot4+160\cdot2+192)+784\cr &=\mathbf{784+560\cdot\sqrt2} \end{aligned}
Did Jane simplify the expression correctly? If not, determine the step at which Jane made a mistake.
Jane made a mistake in step (2). It should be ${6\choose0}=1$. So, the correct result is: $$\left(\sqrt2+2\right)^6=792+560\cdot\sqrt2$$
Jane made a mistake in step (2). It should be ${6\choose6}=0$. So, the correct result is: $$\left(\sqrt2+2\right)^6=720+560\cdot\sqrt2$$
Jane made a mistake in step (2). It should be $\left(\sqrt2\right)^0\cdot64=0$. So, the correct result is: $$\left(\sqrt2+2\right)^6=720+560\cdot\sqrt2$$
Jane made a mistake in step (4) during factoring. The result of this operation should be: $$\sqrt2(12\cdot2+160\cdot2+192)+784$$ and finally it holds: $$\left(\sqrt2+2\right)^6=720+536\cdot\sqrt2$$
\begin{aligned} \left(\sqrt2+2\right)^6 &=\quad{6\choose 0}\left(\sqrt2\right)^6\cdot2^0+{6\choose1}\left(\sqrt2\right)^5\cdot2^1+{6\choose2}\left(\sqrt2\right)^4\cdot2^2+\cr &\ \ \ \ \, +{6\choose3}\left(\sqrt2\right)^3\cdot2^3+{6\choose4}\left(\sqrt2\right)^2\cdot2^4+{6\choose5}\left(\sqrt2\right)^1\cdot2^5+{6\choose6}\left(\sqrt2\right)^0\cdot2^6=\cr &=\color{red}1\color{black}\cdot\left(\sqrt2\right)^6+ 6\cdot\left(\sqrt2\right)^5\cdot2+15\cdot\left(\sqrt2\right)^4\cdot4+20\cdot\left(\sqrt2\right)^3\cdot8+\cr &\ \ \ \ \ +15\cdot\left(\sqrt2\right)^2\cdot16+ 6\cdot\left(\sqrt2\right)^1\cdot32+1\cdot\left(\sqrt2\right)^0\cdot64=\cr &=\left(\sqrt2\right)^6+12\cdot\left(\sqrt2\right)^5+60\cdot\left(\sqrt2\right)^4+160\cdot\left(\sqrt2\right)^3+240\cdot\left(\sqrt2\right)^2+192\cdot\sqrt2+64=\cr &=8+12\cdot\left(\sqrt2\right)^5+60\cdot4+160\cdot\left(\sqrt2\right)^3+240\cdot2+192\cdot\sqrt2+64=\cr &=12\cdot\left(\sqrt2\right)^5+160\cdot\left(\sqrt2\right)^3+192\cdot\sqrt2+792=\cr &=\sqrt2\left(12\cdot\left(\sqrt2\right)^4+160\cdot\left(\sqrt2\right)^2+192\right)+792=\cr &=\sqrt2\cdot(12\cdot4+160\cdot2+192)+792=\cr &=792+560\cdot\sqrt2 \end{aligned}