Ta có (a+2)3-(a+6)(a2+12)+64=a3+6a2+12a+8-a3-12a-6a2-72+64=0(đpcm)
\(\left(a+2^3\right)-\left(a+6\right).\left(a^2+12\right)+64=0\)
\(\Leftrightarrow\left(a+8\right)-\left(a^3+6a^2+12a+72\right)=-64\)
\(\Leftrightarrow\left(a^3+6a^2+12a+72\right)-\left(a+8\right)=64\)
\(\Leftrightarrow a^3+6a^2+11a+64=64\)
\(\Leftrightarrow a^3+6a^2+11a^2=0\)
\(\Leftrightarrow a.\left(a^2+6a+11\right)=0\)
\(\Leftrightarrow a.\left[\left(a^2+2.a.3+9\right)+2\right]=0\)
\(\Leftrightarrow a.\left[\left(a+3\right)^2+2\right]=0\)
\(\Leftrightarrow\left[{}\begin{matrix}a=0\\\left(a+3\right)^2+2=0\left(\text{Vô lí}\right)\end{matrix}\right.\)
\(\Rightarrow a=0\)
\(\Rightarrow\) Đpcm.
