1-if-s-n-n-n-n-where-are-the-root-of-ax-3-bx-2-cx-d-0-then-show-that-s-4-4abd-4b-2-c-2c-a-3-

Question Number 43665 by peter frank last updated on 13/Sep/18

1) if s_(n ) =α^n +β^n +λ^(n ) where α,β,λ are the root of ax^3 +bx^2 +cx+d=0 then show that s_(4 ) =((4abd+4b^2 c−2c)/a^3 )

$$\left.\mathrm{1}\right)\:{if}\:\:{s}_{{n}\:\:} \:=\alpha^{{n}} +\beta^{{n}} +\lambda^{{n}\:} \:{where}\:\alpha,\beta,\lambda \\ $$$${are}\:{the}\:{root}\:{of}\:{ax}^{\mathrm{3}} +{bx}^{\mathrm{2}} +{cx}+{d}=\mathrm{0} \\ $$$$\:{then}\:\:{show}\:{that}\:{s}_{\mathrm{4}\:} =\frac{\mathrm{4}{abd}+\mathrm{4}{b}^{\mathrm{2}} {c}−\mathrm{2}{c}}{{a}^{\mathrm{3}} } \\ $$

Commented by math1967 last updated on 14/Sep/18

$${more}\:{condition}\:{require},{I}\:{think} \\ $$

Answered by MJS last updated on 15/Sep/18

let′s try an example with given values 2(x−3)(x+5)(x−7)=0 2x^3 −10x^2 −58x+210=0 3^4 +(−5)^4 +7^4 =3107 ((4×2×(−10)×210+4×(−10)^2 (−58)−2(−58))/2^3 )=−4985.5 ⇒ not true

$$\mathrm{let}'\mathrm{s}\:\mathrm{try}\:\mathrm{an}\:\mathrm{example}\:\mathrm{with}\:\mathrm{given}\:\mathrm{values} \\ $$$$\mathrm{2}\left({x}−\mathrm{3}\right)\left({x}+\mathrm{5}\right)\left({x}−\mathrm{7}\right)=\mathrm{0} \\ $$$$\mathrm{2}{x}^{\mathrm{3}} −\mathrm{10}{x}^{\mathrm{2}} −\mathrm{58}{x}+\mathrm{210}=\mathrm{0} \\ $$$$\mathrm{3}^{\mathrm{4}} +\left(−\mathrm{5}\right)^{\mathrm{4}} +\mathrm{7}^{\mathrm{4}} =\mathrm{3107} \\ $$$$\frac{\mathrm{4}×\mathrm{2}×\left(−\mathrm{10}\right)×\mathrm{210}+\mathrm{4}×\left(−\mathrm{10}\right)^{\mathrm{2}} \left(−\mathrm{58}\right)−\mathrm{2}\left(−\mathrm{58}\right)}{\mathrm{2}^{\mathrm{3}} }=−\mathrm{4985}.\mathrm{5} \\ $$$$\Rightarrow\:\mathrm{not}\:\mathrm{true} \\ $$

Commented by math1967 last updated on 15/Sep/18

$${You}\:{are}\:{correct}\:{sir} \\ $$

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