Question
Find $(x+1)^6+(x-1)^6$. Hence or otherwise evaluate $(\sqrt{2}+1)^6+(\sqrt{2}-1)^6$

Answer

$( x +1)^6+( x -1)^6=\left[{ }^6 C_0 x^6+{ }^6 C_1 x^5+{ }^6 C_2 x^4+{ }^6 C_3 x^3+{ }^6 C_4 x^2+{ }^6 C_5 x+{ }^6 C_6\right]$
$+\left[{ }^6 C_0 x^6+{ }^6 C_1 x^5(-1)+{ }^6 C_2 x^4(-1)^2+{ }^6 C_3 x^3(-1)^3+{ }^6 C_4 x^2(-1)^4+{ }^6 C_5 x(-1)^5+{ }^6 C_6(-1)^6\right]$
$=\left[x^6+6 x^5+15 x^4+20 x^3+15 x^2+6 x +1\right]+\left[x^6-6 x^5+15 x^4-20 x^3+15 x^2-6 x+1\right]$
$=2 x^6+30 x^4+30 x^2+2$
$=2\left(x^6+15 x^4+15 x^2+1\right)$
Putting $x=\sqrt{2}$
$(\sqrt{2}+1)^6+(\sqrt{2}-1)^6=2\left[(\sqrt{2})^6+15(\sqrt{2})^4+15(\sqrt{2})^2+1\right]$
$=2[8+15 \times 4+15 \times 2+1]$
$=2[8+60+30+1]$
$=2 \times 99=198$

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