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Question

If (z3 + 1z3) - (z2 + 1z2) + (z+1z) - 1 = (z+1z2α) (z+1z2β) (z+1z2γ) Find the value of α.β.γ.


  1. .

  2. .

  3. .

  4. .


Solution

The correct option is C

.


(z3 + 1z3) - (z2 + 1z2) + (z+1z) - 1 = (z+1z2α) (z+1z2β) (z+1z2γ)---------(1)

Lets assume z+1z = 2x--------(2)

  (z+1z)2 = z2 + 1z2 + 2(z×1z)

   z2 + 1z2 = (2x)2 - 2 = 4x2 - 2 ------------(3)

    (z+1z)3 = z3 + 1z3 + 3(z+1z)

     z3 + 1z3 = (2x)3 - 3 ×(2x) = 8x3 - 6x-----------(4)

Substitute the value of (z+1z), z2 + 1z2 and z3 + 1z3 in equation (1)

(8x36x)(4x22)+(2x)1=(2x2α) (2x - 2β) (2x - 2γ)

 8x3 - 4x2 - 4x + 1 = 8(x - α) (x - β) (x - γ)--------------(5)

From equation 5,we see that  α,β and γ are the roots of the equation 8x3 - 4x2 - 4x + 1 = 0

Product of the roots α,β,γ = 18.

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