Đặt 2011 = a ; 11 = b ; 2000 = c
\(\Rightarrow a=b+c\)
Xét vế phải của đẳng thức ta có: \(\frac{2011^3+11^3}{2011^3+2000^3}=\frac{a^3+b^3}{a^3+c^3}=\frac{\left(a+b\right)\left(a^2-ab+b^2\right)}{\left(a+c\right)\left(a^2-ac+c^2\right)}\)
Thay \(a=b+c\)vào \(a^2-ab+b^2=\left(b+c\right)^2-\left(b+c\right).b+b^2=b^2+bc+c^2\)
Thay \(a=b+c\)vào \(a^2-ac+c^2=\left(b+c\right)^2-\left(b+c\right).c+c^2=b^2+bc+c^2\)
\(\Rightarrow\)\(a^2-ab+b^2=a^2-ac+c^2\)
\(\Rightarrow\) \(\frac{2011^3+11^3}{2011^3+2000^3}=\frac{a^3+b^3}{a^3+c^3}=\frac{\left(a+b\right)\left(a^2-ab+b^2\right)}{\left(a+c\right)\left(a^2-ac+c^2\right)}=\frac{a+b}{a+c}=\frac{2011+11}{2011+2000}\)
Vậy \(\frac{2011^3+11^3}{2011^3+2000^3}=\frac{2011+11}{2011+2000}\left(đpcm\right)\)
