Question

Cho 3 số a,b,c thỏa mãn:

\(\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2=6abc\)

Chứng minh: \(a^3+b^3+c^3=3abc\left(a+b+c+1\right)\)

Answer 1

Lời giải:

Ta có:

$a^3+b^3+c^3-3abc=(a+b)^3-3ab(a+b)+c^3-3abc$

$=(a+b)^3+c^3-3ab(a+b+c)$

$=(a+b+c)[(a+b)^2-c(a+b)+c^2]-3ab(a+b+c)$

$=(a+b+c)[(a+b)^2-c(a+b)+c^2-3ab]=(a+b+c)(a^2+b^2+c^2-ab-bc-ac)$

$=\frac{1}{2}(a+b+c)(2a^2+2b^2+2c^2-2ab-2bc-2ac)$

$=\frac{1}{2}(a+b+c)[(a-b)^2+(b-c)^2+(c-a)^2]$

$=\frac{1}{2}(a+b+c).6abc=3abc(a+b+c)$

$\Rightarrow a^3+b^3+c^3=3abc(a+b+c+1)$ (đpcm)