Answer:
(a) 216x³ -216x² -53x -8
(b) 161/108
(c) 133/72
Explanation:
Given that α, β, and γ are the roots of the cubic 2x³ +5x² -6 = 0, you want ...
(a) a cubic with roots 1/α³, 1/β³, 1/γ³
(b) the value of 1/α⁶ +1/β⁶ + 1/γ⁶
(c) the value of 1/α⁹ +1/β⁹ + 1/γ⁹
General case
Given the monic cubic x³ +ax² +bx +c = 0, we know that a, b, c are related to the roots by ...
- a = -α -β -γ
- b = αβ +αγ +βγ
- c = -αβγ
Then if 1/α³, 1/β³, 1/γ³ are the roots of x³ +a'x² +b'x +c' = 0, we can show algebraically that ...
a' = (b³ - 3 a b c + 3 c²)/c³
b' = (a³ - 3 a b + 3 c)/c³
c' = 1/c³
We can also show that ...
α² +β² +γ² = a² -2b
α³ +β³ +γ³ = 3(ab -c) -a³
(a) Inverse cubes
Using the above relations we find the coefficients of the required cubic can be found from a = 5/2, b = 0, c = -3:
a' = (b³ - 3 a b c + 3 c²)/c³ = -1
b' = (a³ - 3 a b + 3 c)/c³ = -53/216
c' = 1/c³ = -1/27
For the next parts, we will use this version of the desired cubic:
x³ -x² -(53/216)x -1/27 = 0
We can write this with integer coefficients as ...
216x³ -216x² -53x -8 = 0 . . . . . . equation of the required cubic
(b) Sum of -6th powers
The cubic we just found has roots 1/α³, 1/β³, 1/γ³, and we want the value of the sum of the squares of these roots: 1/α⁶ +1/β⁶ + 1/γ⁶.
For the purpose we can use the relation α² +β² +γ² = a² -2b, where the coefficients are the a' and b' we found above
1/α⁶ +1/β⁶ + 1/γ⁶ = (-1)² -2(-53/216) = 1 +106/216
1/α⁶ +1/β⁶ + 1/γ⁶ = 161/108
(c) Sum of -9th powers
As above, we need to find the sum of the cubes of the roots of the new cubic. This is given by the relation α³ +β³ +γ³ = 3(ab -c) -a³.
1/α⁹ +1/β⁹ + 1/γ⁹ = 3(-1(-53/216 -(-1/27)) -(-1)³
1/α⁹ +1/β⁹ + 1/γ⁹ = 133/72
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Additional comment
The first attachment shows the calculations using the formulas we found. The second attachment shows a machine solution to this problem. Note that the original equation has one real root and two complex roots.
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