Question

Cho a+b+c=abc CMR:

\(a\left(b^2-1\right)\left(c^2-1\right)+b\left(a^2-1\right)\left(c^2-1\right)+c\left(a^2-1\right)\left(b^2-1\right)=4abc\)

Answer 1

Lời giải:

Ta có:
\(a(b^2-1)(c^2-1)+b(a^2-1)(c^2-1)+c(a^2-1)(b^2-1)\)

\(=a(b^2c^2-b^2-c^2+1)+b(a^2c^2-a^2-c^2+1)+c(a^2b^2-a^2-b^2+1)\)

\(=(ab^2c^2+ba^2c^2+ca^2b^2)+(a+b+c)-[a(b^2+c^2)+b(a^2+c^2)+c(a^2+b^2)]\)

\(=abc(ab+bc+ac)+abc-[ab(a+b)+bc(b+c)+ca(c+a)]\)

\(=abc(ab+bc+ca)+4abc-[ab(a+b+c)+bc(b+c+a)+ca(c+a+b)]\)

\(=abc(ab+bc+ca)+4abc-(a+b+c)(ab+bc+ac)\)

\(=abc(ab+bc+ca)+4abc-abc(ab+bc+ac)=4abc\)

Ta có đpcm.