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As α, β and γ are the roots of the equation, (x - α)(x - β)(x - γ) = 0http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics - Further (9231)/9231_s10_qp_12.pdf
Question 6. There has to be a formula for S6 right?
Please help theorem you use for these kind of summatories, in which you have to cube S2 and all that! I'm very lost.
Expanding the above, x³ - x²(α + β + γ) + x(αβ + βγ + γα) - αβγ = 0 ------- (1)
But given is: x³ + x - 1 = 0 --------- (2)
Comparing the corresponding coefficients of (1) & (2):
(α + β + γ) = 0; (αβ + βγ + γα) = 1 and αβγ = 1 --------------- (3)
As x = √y; x² = y; ==> y = α² or β² or γ² [Since, α, β and γ are the values of x as given]
==> (y - α²)(y - β²)(y - γ²) = 0
Expanding, y³ - (α² + β² + γ²)y² + (α²β² + β²γ² + α²γ²)y - α²*β²*γ² = 0 ------ (4)
By identity, (α + β + γ)² = (α² + β² + γ²) + 2(αβ + βγ + γα)
Substituting the values from (3) and simplifying, (α² + β² + γ²) = -2
(αβ + βγ + γα)² = (α²β² + β²γ² + α²γ²) + 2(αβγ)(αβ + βγ + γα)
Substituting the values from (3) and simplifying, (α²β² + β²γ² + α²γ²) = 1
and α²*β²*γ² = (αβγ)² = 1
Substituting all these values in (4):
y³ + 2y² + y - 1 = 0 [Proved]
Already it is shown that: (α² + β² + γ²) = -2; so S₂ = -2
Squaring the above, we can get S₄ = 2
For part (ii) you may try yourself in similar to the above. If you don't get, knock me anytime.