Question 1205175: If 𝛼, 𝛽, 𝛾 (where 𝛼, 𝛽, 𝛾 ≠ 0) are the roots of the equation 𝑥
3 + 𝑝𝑥^2 + 𝑞𝑥 + 𝑟 = 0, where 𝑝, 𝑞
and 𝑟 (≠ 0) are real numbers, express the following in terms of 𝑝, 𝑞 and 𝑟:
1/𝛼^3 + 1/𝛽^3 + 1/𝛾^3
Answer by ikleyn(52932)   (Show Source):
You can put this solution on YOUR website! .
If 𝛼, 𝛽, 𝛾 (where 𝛼, 𝛽, 𝛾 ≠ 0) are the roots of the equation 𝑥^3 + 𝑝𝑥^2 + 𝑞𝑥 + 𝑟 = 0,
where 𝑝, 𝑞 and 𝑟 (≠ 0) are real numbers, express the following in terms of 𝑝, 𝑞 and 𝑟:
1/𝛼^3 + 1/𝛽^3 + 1/𝛾^3
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For simplicity of writing, I will replace  ,  and  by "a", "b" and "c".
So, we are given an equation  = 0, where p, q and r (=/=0) are real numbers,
with the roots a, b and c.
They want we find  +  +  .
Step by step solution
(a) First, notice that if "a" is the solution to polynomial equation = 0, then
 = 0. (1)
Since r =/= 0, the root "a" is also not zero, a =/= 0. In equation (1), divide both sides by  .
You will get then
 = 0.
It means that  is the root of the cubic polynomial equation
 = 0. (2)
Similarly, if "a", "b" and "c" are the roots to equation (1), then ,  and  are the roots
of equation (2).
(b) OK. It means that if "a", "b" and "c" are the roots of equation (1), = 0,
they want we calculate  , where d, e, and f are the roots of equation (2), = 0.
(c) Due to Vieta's theorem, if d, e and f are the roots of equation (2), then
d + e + f =  , d*e + d*f + e*f =  , d*e*f =  . (3)
(d) For any real numbers d, e, f, the following identity is valid
 = + 3*(d+e+f)*(de + df + ef) - 3def. (4)
It can be checked / proved by direct calculation.
(e) Now, substitute expressions (3) into (4). You will get then
 = +  -  .
It implies = +  -  , or
=  +  -  .
(f) Thus the problem is just solved, and the ANSWER is:
if a, b and c are the roots of equation (1), then + + =  + - .
ANSWER. If a, b and c are the roots of equation = 0,
then + + = + - .
Solved.
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