Show that all roots of `(x+1)^6+(x-1)^6=0` are given by -icot`((2k+1)n)/12`where k=0,1,2,3,4,5.

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#### Solution

`(x+1)^6+(x-1)^6=0`

∴ `(x+1)^6=(-x-1)^6`

∴ `(x+1)^6/(x-1)^6=-1`

∴ `((x+1)/(x-1))^6=e^(ipi)` `{∵ e^(ipi)=cospi+isinpi=-1+i(0)=-1}` (Principal value)

∴` ((x+1)/(x-1)^6=e^(i(pi+2kpi))` , k=0,1,2,3,4,5

∴ `(x+1)/(x-1)=e(ipi(1+2k)/6)`

Let `2θ= (pi(1+2k))/6`

∴ From (1) & (2),` (x+1)/(x-1)=e^(i2θ)`

∴ By Componendo – Dividendo,`( (x+1)+(x-1))/((x+1)-(x-1))=(e^12θ+1)/(e^12θ-1)`

∴ `2x/2= (e^(iθ)[e^(iθ)+e^(-iθ)])/(e^(iθ)[e^(iθ)-e^(-iθ)])`

∴ `x=(2cosθ)/(2isinθ)` {∵`sinθ=(e^(iθ)-e^(-iθ))/(2i) and cos θ= (e^(iθ)+e^(-iθ))/(2) }`

∴` x=1/i cotθ `

∴ `x=-icot ((2k+1)n)/12`(From 2) where k = 0, 1, 2, 3, 4, 5

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Concept: Powers and Roots of Trigonometric Functions

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