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Question Number 198267 by Fridunatjan08 last updated on 16/Oct/23
Find the real values of n: n^6 −n^3 =2
$${Find}\:{the}\:{real}\:{values}\:{of}\:{n}:\:{n}^{\mathrm{6}} −{n}^{\mathrm{3}} =\mathrm{2} \\ $$
Answered by Rasheed.Sindhi last updated on 16/Oct/23
n^6 −n^3 =2 n^6 −n^3 −2=0 n^6 +2n^3 −3n^3 +1−3=0 n^6 +2n^3 +1−3n^3 −3=0 (n^3 +1)^2 −3(n^3 +1)=0 (n^3 +1)(n^3 +1−3)=0 n^3 +1=0 ∣ n^3 −2=0 n^3 =−1 ∣ n^3 =2 n=−1, (2)^(1/3)
$${n}^{\mathrm{6}} −{n}^{\mathrm{3}} =\mathrm{2} \\ $$$${n}^{\mathrm{6}} −{n}^{\mathrm{3}} −\mathrm{2}=\mathrm{0} \\ $$$${n}^{\mathrm{6}} +\mathrm{2}{n}^{\mathrm{3}} −\mathrm{3}{n}^{\mathrm{3}} +\mathrm{1}−\mathrm{3}=\mathrm{0} \\ $$$${n}^{\mathrm{6}} +\mathrm{2}{n}^{\mathrm{3}} +\mathrm{1}−\mathrm{3}{n}^{\mathrm{3}} −\mathrm{3}=\mathrm{0} \\ $$$$\left({n}^{\mathrm{3}} +\mathrm{1}\right)^{\mathrm{2}} −\mathrm{3}\left({n}^{\mathrm{3}} +\mathrm{1}\right)=\mathrm{0} \\ $$$$\left({n}^{\mathrm{3}} +\mathrm{1}\right)\left({n}^{\mathrm{3}} +\mathrm{1}−\mathrm{3}\right)=\mathrm{0} \\ $$$${n}^{\mathrm{3}} +\mathrm{1}=\mathrm{0}\:\:\mid\:\:{n}^{\mathrm{3}} −\mathrm{2}=\mathrm{0} \\ $$$${n}^{\mathrm{3}} =−\mathrm{1}\:\mid\:\:{n}^{\mathrm{3}} =\mathrm{2} \\ $$$${n}=−\mathrm{1},\:\sqrt[{\mathrm{3}}]{\mathrm{2}}\: \\ $$
Answered by Rasheed.Sindhi last updated on 16/Oct/23
n^6 −n^3 −2=0 n^6 −2n^3 +n^3 −2=0 n^3 (n^3 −2)+(n^3 −2)=0 (n^3 −2)(n^3 +1)=0 n^3 =−1 , 2 n=−1, (2)^(1/3)
$${n}^{\mathrm{6}} −{n}^{\mathrm{3}} −\mathrm{2}=\mathrm{0} \\ $$$${n}^{\mathrm{6}} −\mathrm{2}{n}^{\mathrm{3}} +{n}^{\mathrm{3}} −\mathrm{2}=\mathrm{0} \\ $$$${n}^{\mathrm{3}} \left({n}^{\mathrm{3}} −\mathrm{2}\right)+\left({n}^{\mathrm{3}} −\mathrm{2}\right)=\mathrm{0} \\ $$$$\left({n}^{\mathrm{3}} −\mathrm{2}\right)\left({n}^{\mathrm{3}} +\mathrm{1}\right)=\mathrm{0} \\ $$$${n}^{\mathrm{3}} =−\mathrm{1}\:,\:\mathrm{2} \\ $$$${n}=−\mathrm{1},\:\sqrt[{\mathrm{3}}]{\mathrm{2}}\: \\ $$
Answered by Frix last updated on 16/Oct/23
n=(t)^(1/3) t^2 −t=2 t=−1∨t=2 n=−1∨n=(2)^(1/3)
$${n}=\sqrt[{\mathrm{3}}]{{t}} \\ $$$${t}^{\mathrm{2}} −{t}=\mathrm{2} \\ $$$${t}=−\mathrm{1}\vee{t}=\mathrm{2} \\ $$$${n}=−\mathrm{1}\vee{n}=\sqrt[{\mathrm{3}}]{\mathrm{2}} \\ $$

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