Question

[Ngô Phúc Dương]

Cho a, b, c thỏa mãn (a + b + c)(ab + bc + ac) = 2018 và abc = 2018. Tính giá trị của biểu thức P = (b^2.c + 2018)(a^2.b + 2018)(c^2.a + 2018)

Answer 1

[Luân Đào]

\(P=\left(b^2c+abc\right)\left(a^2b+abc\right)\left(c^2a+abc\right)\)

\(=bc\left(a+b\right)\cdot ab\left(c+a\right)\cdot ca\left(b+c\right)\)

\(=\left(abc\right)^2\left(a+b\right)\left(b+c\right)\left(c+a\right)\)

Lại có:

\(\left(a+b+c\right)\left(ab+bc+ca\right)-abc=0\)

\(\Leftrightarrow\left(a^2b+abc+a^2c\right)+\left(ab^2+b^2c+abc\right)+\left(bc^2+c^2a+abc\right)-abc=0\)

\(\Leftrightarrow a^2b+ca^2+ab^2+2abc+ac^2+b^2c+bc^2=0\)

\(\Leftrightarrow a^2\left(b+c\right)+a\left(b^2+2bc+c^2\right)+bc\left(b+c\right)=0\)

\(\Leftrightarrow a^2\left(b+c\right)+a\left(b+c\right)^2+bc\left(b+c\right)=0\)

\(\Leftrightarrow\left(b+c\right)\left(a^2+ab+ca+bc\right)=0\)

\(\Leftrightarrow\left(b+c\right)\left[a\left(a+b\right)+c\left(a+b\right)\right]=0\)

\(\Leftrightarrow\left(b+c\right)\left(a+b\right)\left(c+a\right)=0\)

\(\Rightarrow P=0\)