Question

If α, β, γ are roots of equation x3

+px + q = 0. Then show that

(α7

+ β7

+ γ7

) = (α2

+ β2

+ γ2

) (α5

+ β5

+ γ5

)

Answer 1

I faced same kind of question. Look at bit 5
Let α; β; γ be the roots of x3 - 3x2 + 1.
1 Find a polynomial whose roots are α + 3, β + 3, and γ + 3.
2 Find a polynomial whose roots are α+3 1 , β+3 1 , and γ+3 1 .
3 Compute α+3 1 + β+3 1 + γ+3 1 .
4 Find a polynomial whose roots are α2, β2, and γ2.
5 Find a recurrence relation for xn = αn + βn + γn, and use it to
compute α5 + β5 + γ5.

Answer:
1 To increment all roots by 3, substitute x - 3 for x. This yields
x3 - 12x2 + 45x - 53.
2 Reversing the coefficients to get 1 - 12x + 45x2 - 53x3 yields
a polynomial whose roots are reciprocals of the polynomial
above. (Do you see why?)
3 This is just the sum of the roots of the polynomial above: 45 53.
4 Observe that x3 + 3x2 - 1 has roots -α, -β, and -γ.
Therefore (x3 - 3x2 + 1)(x3 + 3x2 - 1) has roots ±α; ±β; ±γ
and factors as (x2 - α2)(x2 - β2)(x2 - γ2). Replacing x2 by
x yields our answer: x3 - 9x2 + 6x - 1.
5 The recurrence is xn = 3xn-1 - xn-3. We have x0 = 3,
x1 = 3, and x2 = 9, so x3 = 24, x4 = 69, and x5 = 198

HOPE IT HELPS!!!!