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Question Number 165329 by mnjuly1970 last updated on 30/Jan/22
prove ( n∈ N ) 3(n+1) ∣ n^( 3) + (n+1)^( 3) + (n+2 )^( 3)
prove(n∈N)3(n+1)∣n3+(n+1)3+(n+2)3
Answered by Rasheed.Sindhi last updated on 30/Jan/22
3(n+1) ∣ n^( 3) + (n+1)^( 3) +(n+2 )^( 3) _(−) ∵ n^( 3) + (n+1)^( 3) +(n+2 )^( 3) =(n+1−1)^3 + (n+1)^( 3) +(n+1+1 )^( 3) =(n+1)^3 −1−3(n+1)(n+1−1) +(n+1)^3 +(n+1)^3 +1+3(n+1)(n+1+1) =3(n+1)^3 −3n(n+1)+3(n+1)(n+2) =3(n+1){(n+1)^2 −3n+n+2} ∴ 3(n+1) ∣ n^( 3) + (n+1)^( 3) +(n+2 )^( 3)
3(n+1)∣n3+(n+1)3+(n+2)3−∵n3+(n+1)3+(n+2)3=(n+1−1)3+(n+1)3+(n+1+1)3=(n+1)3−1−3(n+1)(n+1−1)+(n+1)3+(n+1)3+1+3(n+1)(n+1+1)=3(n+1)3−3n(n+1)+3(n+1)(n+2)=3(n+1){(n+1)2−3n+n+2}∴3(n+1)∣n3+(n+1)3+(n+2)3
Commented by mnjuly1970 last updated on 30/Jan/22
very nice solution sir ...
verynicesolutionsir…
Answered by Rasheed.Sindhi last updated on 31/Jan/22
AnOther method 3(n+1) ∣ n^3 + (n+1)^3 + (n+2 )^3 _(−) ∵ n^3 + (n+1)^3 + (n+2 )^3 =a^3 +b^3 +c^3 −3abc+3abc ; { ((a=n)),((b=n+1)),((c=n+2)) :} =(a+b+c)(a^2 +b^2 +c^2 −ab−bc−ca)+3abc =(n+n+1+n+2)( ∼ )+3n(n+1)(n+2) =3(n+1)( ∼ )+3n(n+1)(n+2) =3(n+1){( ∼ )+n(n+2)} ∴ 3(n+1) ∣ n^3 + (n+1)^3 +(n+2 )^3
AnOthermethod3(n+1)∣n3+(n+1)3+(n+2)3−∵n3+(n+1)3+(n+2)3=a3+b3+c3−3abc+3abc;{a=nb=n+1c=n+2=(a+b+c)(a2+b2+c2−ab−bc−ca)+3abc=(n+n+1+n+2)(∼)+3n(n+1)(n+2)=3(n+1)(∼)+3n(n+1)(n+2)=3(n+1){(∼)+n(n+2)}∴3(n+1)∣n3+(n+1)3+(n+2)3
Commented by mnjuly1970 last updated on 31/Jan/22
thank you so much sir Rasheed
thankyousomuchsirRasheed

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