Find (x + 1)6 + (x – 1)6. Hence or otherwise evaluate.

Using Binomial Theorem, the expressions, $(x+1)^{6}$ and $(x-1)^{6}$, can be expanded as

$(x+1)^{6}={ }^{6} \mathrm{C}_{0} x^{6}+{ }^{6} \mathrm{C}_{1} x^{5}+{ }^{6} \mathrm{C}_{2} x^{4}+{ }^{6} \mathrm{C}_{3} x^{3}+{ }^{6} \mathrm{C}_{4} x^{2}+{ }^{6} \mathrm{C}_{5} x+{ }^{6} \mathrm{C}_{6}$

$(x-1)^{6}={ }^{6} \mathrm{C}_{0} x^{6}-{ }^{6} \mathrm{C}_{1} x^{5}+{ }^{6} \mathrm{C}_{2} x^{4}-{ }^{6} \mathrm{C}_{3} x^{3}+{ }^{6} \mathrm{C}_{4} x^{2}-{ }^{6} \mathrm{C}_{5} x+{ }^{6} \mathrm{C}_{6}$

$\begin{aligned} \therefore(x+1)^{6}+(x-1)^{6} &=2\left[{ }^{6} \mathrm{C}_{0} x^{6}+{ }^{6} \mathrm{C}_{2} x^{4}+{ }^{6} C_{4} x^{2}+{ }^{6} C_{6}\right] \\ &=2\left[x^{6}+15 x^{4}+15 x^{2}+1\right] \end{aligned}$

By putting $x=\sqrt{2}$, we obtain

$(\sqrt{2}+1)^{6}+(\sqrt{2}-1)^{6}=2\left[(\sqrt{2})^{6}+15(\sqrt{2})^{4}+15(\sqrt{2})^{2}+1\right]$

$=2(8+15 \times 4+15 \times 2+1)$

$=2(8+60+30+1)$

$=2(99)=198$