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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
Findtherealvaluesofn:n6−n3=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)
n6−n3=2n6−n3−2=0n6+2n3−3n3+1−3=0n6+2n3+1−3n3−3=0(n3+1)2−3(n3+1)=0(n3+1)(n3+1−3)=0n3+1=0∣n3−2=0n3=−1∣n3=2n=−1,23
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)
n6−n3−2=0n6−2n3+n3−2=0n3(n3−2)+(n3−2)=0(n3−2)(n3+1)=0n3=−1,2n=−1,23
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=t3t2−t=2t=−1∨t=2n=−1∨n=23

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