divide n^3-3n^2+5n-3 by n^2-2
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If 3 is a multiple of n then3|n3+3n2+5n+3
Let gcd (n , 3) = 1. Then by Fermat’s little theorem
n2≡1(mod3)
⟹n3≡n(mod3)
⟹n3≡n−6n(mod3)
⟹n3≡−5n(mod3)
⟹n3+5n≡0(mod3)
⟹n3+3n2+5n+3≡0(mod3)
∴3|n3+3n2+5n+3
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