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Question Number 173152 by AgniMath last updated on 07/Jul/22

If x = 2 + 2^(2/3) + 2^(1/3) then prove that x^3 − 6x^2 + 6x − 2 = 0.

$$\mathrm{If}\:{x}\:=\:\mathrm{2}\:+\:\mathrm{2}^{\frac{\mathrm{2}}{\mathrm{3}}} \:+\:\mathrm{2}^{\frac{\mathrm{1}}{\mathrm{3}}} \:\mathrm{then}\:\mathrm{prove}\:\mathrm{that}\: \\ $$$${x}^{\mathrm{3}} \:−\:\mathrm{6}{x}^{\mathrm{2}} \:+\:\mathrm{6}{x}\:−\:\mathrm{2}\:=\:\mathrm{0}. \\ $$

Answered by Frix last updated on 07/Jul/22

let x=y^3 +y^2 +y y^9 +3y^8 +6y^7 +y^6 −6y^5 −15y^4 −5y^3 +6y−2=0 y=2^(1/3) 8+2^(2/3) 12+2^(1/3) 24+4−2^(2/3) 12−2^(1/3) 30−10+2^(1/3) 6−2=0 ir′s easy to see this is true.

$$\mathrm{let}\:{x}={y}^{\mathrm{3}} +{y}^{\mathrm{2}} +{y} \\ $$$${y}^{\mathrm{9}} +\mathrm{3}{y}^{\mathrm{8}} +\mathrm{6}{y}^{\mathrm{7}} +{y}^{\mathrm{6}} −\mathrm{6}{y}^{\mathrm{5}} −\mathrm{15}{y}^{\mathrm{4}} −\mathrm{5}{y}^{\mathrm{3}} +\mathrm{6}{y}−\mathrm{2}=\mathrm{0} \\ $$$${y}=\mathrm{2}^{\mathrm{1}/\mathrm{3}} \\ $$$$\mathrm{8}+\mathrm{2}^{\mathrm{2}/\mathrm{3}} \mathrm{12}+\mathrm{2}^{\mathrm{1}/\mathrm{3}} \mathrm{24}+\mathrm{4}−\mathrm{2}^{\mathrm{2}/\mathrm{3}} \mathrm{12}−\mathrm{2}^{\mathrm{1}/\mathrm{3}} \mathrm{30}−\mathrm{10}+\mathrm{2}^{\mathrm{1}/\mathrm{3}} \mathrm{6}−\mathrm{2}=\mathrm{0} \\ $$$$\mathrm{ir}'\mathrm{s}\:\mathrm{easy}\:\mathrm{to}\:\mathrm{see}\:\mathrm{this}\:\mathrm{is}\:\mathrm{true}. \\ $$

Answered by som(math1967) last updated on 07/Jul/22

x−2=2^(2/3) +2^(1/3) (x−2)^3 =(2^(2/3) +2^(1/3) )^3 ⇒x^3 −6x^2 +12x−8=(2^(2/3) )^3 +(2^(1/3) )^3 +3.2^(2/3) .2^(1/3) (2^(2/3) +2^(1/3) ) ⇒x^3 −6x^2 +12x−8=4+2+3.2(x−2) [∵ (x−2)=2^(2/3) +2^(1/3) ] ⇒x^3 −6x^2 +12x−8=6+6x−12 ⇒x^3 −6x^2 +12x−6x−8+6=0 ∴x^3 −6x^2 +6x−2=0

$${x}−\mathrm{2}=\mathrm{2}^{\frac{\mathrm{2}}{\mathrm{3}}} +\mathrm{2}^{\frac{\mathrm{1}}{\mathrm{3}}} \\ $$$$\left({x}−\mathrm{2}\right)^{\mathrm{3}} =\left(\mathrm{2}^{\frac{\mathrm{2}}{\mathrm{3}}} +\mathrm{2}^{\frac{\mathrm{1}}{\mathrm{3}}} \right)^{\mathrm{3}} \\ $$$$\Rightarrow{x}^{\mathrm{3}} −\mathrm{6}{x}^{\mathrm{2}} +\mathrm{12}{x}−\mathrm{8}=\left(\mathrm{2}^{\frac{\mathrm{2}}{\mathrm{3}}} \right)^{\mathrm{3}} +\left(\mathrm{2}^{\frac{\mathrm{1}}{\mathrm{3}}} \right)^{\mathrm{3}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:+\mathrm{3}.\mathrm{2}^{\frac{\mathrm{2}}{\mathrm{3}}} .\mathrm{2}^{\frac{\mathrm{1}}{\mathrm{3}}} \left(\mathrm{2}^{\frac{\mathrm{2}}{\mathrm{3}}} +\mathrm{2}^{\frac{\mathrm{1}}{\mathrm{3}}} \right) \\ $$$$\Rightarrow{x}^{\mathrm{3}} −\mathrm{6}{x}^{\mathrm{2}} +\mathrm{12}{x}−\mathrm{8}=\mathrm{4}+\mathrm{2}+\mathrm{3}.\mathrm{2}\left({x}−\mathrm{2}\right) \\ $$$$\left[\because\:\left({x}−\mathrm{2}\right)=\mathrm{2}^{\frac{\mathrm{2}}{\mathrm{3}}} +\mathrm{2}^{\frac{\mathrm{1}}{\mathrm{3}}} \right] \\ $$$$\Rightarrow{x}^{\mathrm{3}} −\mathrm{6}{x}^{\mathrm{2}} +\mathrm{12}{x}−\mathrm{8}=\mathrm{6}+\mathrm{6}{x}−\mathrm{12} \\ $$$$\Rightarrow{x}^{\mathrm{3}} −\mathrm{6}{x}^{\mathrm{2}} +\mathrm{12}{x}−\mathrm{6}{x}−\mathrm{8}+\mathrm{6}=\mathrm{0} \\ $$$$\therefore{x}^{\mathrm{3}} −\mathrm{6}{x}^{\mathrm{2}} +\mathrm{6}{x}−\mathrm{2}=\mathrm{0} \\ $$

Commented by Tawa11 last updated on 11/Jul/22

Great sir

$$\mathrm{Great}\:\mathrm{sir} \\ $$

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