Question

Find the square root of (x+1)^6+1/(x+1 )^6+2

Answer 1

Answer:

Toolbox:

   (1+x)n=nC0+nC1x+nC2x2+....nC1Xr+....+nCnxn

(x+1)6=x6+6C1x5.1+6C2.x4.12+6C3.x3.13+6C4.x2.14+6C5.x.15+6C6.x4.16

⇒x6+6x5+15x4+20x3+15x2+6x+1

-----(1)

(x−1)6=x6+6C1x5.(−1)+6C2.x4.(−1)2+6C3.x3.(−1)3+6C4.x2.(−1)4+6C5.x.(−1)5+6C6.x0.(−1)6

⇒x6−6x5+15x4−20x3+15x2−6x+1

-----(2)

Adding (1) and (2)

(x+1)6+(x−1)6=2[x6+15x4+15x2+1]

Putting 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

Step-by-step explanation:

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