Fewpal
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Find (x + 1)6 + (x – 1)6. Hence or otherwise evaluate ( √ 2 + 1)6 + ( √ 2 – 1)6.

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Solution:

(x+1)= x6+6C1 x5.1+ 6C2 x4.12+6C3 x3.13+6C4 x2.14+6C5 x1 .15+6C6.x0.16

⇒x6+6x5+15x4+20x3+15x2+6x+1 -----(1)

(x−1)6=x6+6C1 x5.(−1)+6C2 x4.(−1)2+6C3 x3.(−1)3+6C4 x2.(−1)4+6C5 x1.(−1)5+6C6.x0.(−1)6

⇒x6−6x5+15x4−20x3+15x2−6x+1 -----(2)

Adding (1) and (2)

(x+1)6+(x−1)= 2[x6+15x4+15x2+1]

Putting x=2–√x=2

(2–√+1)6+(2–√−1)6 = 2[(2–√)6+15(2–√)4+15(2–√)2+1]

⇒2[8+60+30+1]

⇒2[99]

⇒198

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