Question-220810

Question Number 220810 by Rojarani last updated on 19/May/25

Commented by Ghisom last updated on 19/May/25

without further information we can let a=b=c=k ⇒ k=((2^(1/3) −1)/(27)) ⇒ a+b+c=3k=((2^(1/3) −1)/9)

$$\mathrm{without}\:\mathrm{further}\:\mathrm{information}\:\mathrm{we}\:\mathrm{can}\:\mathrm{let} \\ $$$${a}={b}={c}={k}\:\Rightarrow \\ $$$${k}=\frac{\mathrm{2}^{\mathrm{1}/\mathrm{3}} −\mathrm{1}}{\mathrm{27}}\:\Rightarrow \\ $$$${a}+{b}+{c}=\mathrm{3}{k}=\frac{\mathrm{2}^{\mathrm{1}/\mathrm{3}} −\mathrm{1}}{\mathrm{9}} \\ $$

Commented by mr W last updated on 19/May/25

$${a},\:{b},\:{c}\:\in\:{Z}\:{or} \\ $$$${a},\:{b},\:{c}\:\in\:{Q}\:? \\ $$

Commented by Rojarani last updated on 19/May/25

$$\:{Sir},\:\left({a},{b},{c}\right)\in{Q} \\ $$

Answered by Frix last updated on 19/May/25

a>b>c: a=(4/9)∧b=(1/9)∧c=−(2/9) a+b+c=(1/3) Test: ((4/9))^(1/3) +((1/9))^(1/3) −((2/9))^(1/3) =((2^(2/3) −2^(1/3) +1)/3^(2/3) ) ((2^(2/3) −2^(1/3) +1)/3^(2/3) ) =^? (2^(1/3) −1)^(1/3) 2^(2/3) −2^(1/3) +1 =^? 3^(2/3) (2^(1/3) −1)^(1/3) (2^(2/3) −2^(1/3) +1)^3 =^? 9(2^(1/3) −1) True

$${a}>{b}>{c}: \\ $$$${a}=\frac{\mathrm{4}}{\mathrm{9}}\wedge{b}=\frac{\mathrm{1}}{\mathrm{9}}\wedge{c}=−\frac{\mathrm{2}}{\mathrm{9}} \\ $$$${a}+{b}+{c}=\frac{\mathrm{1}}{\mathrm{3}} \\ $$$$\mathrm{Test}: \\ $$$$\left(\frac{\mathrm{4}}{\mathrm{9}}\right)^{\frac{\mathrm{1}}{\mathrm{3}}} +\left(\frac{\mathrm{1}}{\mathrm{9}}\right)^{\frac{\mathrm{1}}{\mathrm{3}}} −\left(\frac{\mathrm{2}}{\mathrm{9}}\right)^{\frac{\mathrm{1}}{\mathrm{3}}} =\frac{\mathrm{2}^{\frac{\mathrm{2}}{\mathrm{3}}} −\mathrm{2}^{\frac{\mathrm{1}}{\mathrm{3}}} +\mathrm{1}}{\mathrm{3}^{\frac{\mathrm{2}}{\mathrm{3}}} } \\ $$$$\frac{\mathrm{2}^{\frac{\mathrm{2}}{\mathrm{3}}} −\mathrm{2}^{\frac{\mathrm{1}}{\mathrm{3}}} +\mathrm{1}}{\mathrm{3}^{\frac{\mathrm{2}}{\mathrm{3}}} }\:\overset{?} {=}\:\left(\mathrm{2}^{\frac{\mathrm{1}}{\mathrm{3}}} −\mathrm{1}\right)^{\frac{\mathrm{1}}{\mathrm{3}}} \\ $$$$\mathrm{2}^{\frac{\mathrm{2}}{\mathrm{3}}} −\mathrm{2}^{\frac{\mathrm{1}}{\mathrm{3}}} +\mathrm{1}\:\overset{?} {=}\:\mathrm{3}^{\frac{\mathrm{2}}{\mathrm{3}}} \left(\mathrm{2}^{\frac{\mathrm{1}}{\mathrm{3}}} −\mathrm{1}\right)^{\frac{\mathrm{1}}{\mathrm{3}}} \\ $$$$\left(\mathrm{2}^{\frac{\mathrm{2}}{\mathrm{3}}} −\mathrm{2}^{\frac{\mathrm{1}}{\mathrm{3}}} +\mathrm{1}\right)^{\mathrm{3}} \:\overset{?} {=}\:\mathrm{9}\left(\mathrm{2}^{\frac{\mathrm{1}}{\mathrm{3}}} −\mathrm{1}\right) \\ $$$$\mathrm{True} \\ $$

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