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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.