Question:
Find $(x+1)^{6}+(x-1)^{6}$. Hence, or otherwise evaluate $(\sqrt{2}+1)^{6}+\sqrt{2}-1^{6}$.
Solution:
The expression $(x+1)^{6}+(x-1)^{6}$ can be written as
$(x+1)^{6}+(x-1)^{6}$
$=2\left[{ }^{6} C_{0} x^{6}+{ }^{6} C_{2} x^{4}+{ }^{6} C_{4} x^{2}+{ }^{6} C_{6} x^{0}\right]$
$=2\left[x^{6}+15 x^{4}+15 x^{2}+1\right]$
By taking $x=\sqrt{2}$, we get:
$(\sqrt{2}+1)^{6}+(\sqrt{2}-1)^{6}=2\left[(\sqrt{2})^{6}+15(\sqrt{2})^{4}+15(\sqrt{2})^{2}+1\right]$
$=2[8+15 \times 4+15 \times 2+1]$
$=2 \times(8+60+30+1)$
$=198$
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