Question

a cho a + b+ c =2019 Chứng minh rằng \(a^3+b^3+c^3⋮\)   3  (a;b;c \(\varepsilonℤ\))

Cho đa thức f(x) = \(x^{99}+x^{88}+x^{77}+...+x^{11}+1\)

                   g(x) = \(x^9+x^8+x^7+...+x+1\)

            Chứng minh rằng f(x)  chia hết cho g(x)

Answer 1

\(\left(a+b+c\right)^3=a^3+b^3+c^3+3\left(a+b\right)\left(b+c\right)\left(c+a\right)\)

\(\Rightarrow\left(a+b+c\right)^3-3\left(a+b\right)\left(b+c\right)\left(c+a\right)=2019^3-3\left(a+b\right)\left(b+c\right)\left(c+a\right)=a^3+b^3+c^3\)

\(2019⋮3\Rightarrow2019^3⋮3\left(1\right)\)

\(3⋮3;a,b,c\in Z\Rightarrow3\left(a+b\right)\left(b+c\right)\left(c+a\right)⋮3\left(2\right)\)

từ (1) và (2) \(\Rightarrow a^3+b^3+c^3⋮3\)