Question

Cho các số nguyên dương a,b,c t/mãn \(\frac{a}{b}=\frac{b}{c}\) CMR:\(\frac{a^3-2b^3+c^3}{a+b+c}\)là số nguyên

Help me plz. ^-^^-^

Answer 1

Từ giả thiết: \(\frac{a}{b}=\frac{b}{c}\Rightarrow ac=b^2\Rightarrow abc=b^3\)

Ta có: \(\frac{a^3-2b^3+c^3}{a+b+c}=\frac{a^3+b^3+c^3-3c^3}{a+b+c}=\frac{a^3+b^3+c^3-3abc}{a+b+c}\)

Xét: \(a^3+b^3+c^3=\left(a+b\right)^3+c^3-3ab\left(a+b\right)-3abc\)

\(=\left(a+b+c\right)\left[\left(a+b\right)^2-c\left(a+b\right)+c^2\right]-3ab\left(a+b+c\right)\)

\(=\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ac\right)\)

\(\Rightarrow\frac{a^3-2b^3+c^3}{a+b+c}=a^2+b^2+c^2-ab-bc-ac\) là 1 số nguyên (đpcm)