Ta có: \(a^3-3a^2+8a=9\)
\(\Leftrightarrow\left(a^3-3a^2+3a-1\right)+5a-8=0\)
\(\Leftrightarrow\left(a-1\right)^3+5a-8=0\)
Lại có: \(b^3-6b^2+17b=15\)
\(\Leftrightarrow\left(b^3-6b^2+12b-8\right)+5b-7=0\)
\(\Leftrightarrow\left(b-2\right)^3+5b-7=0\)
Cộng 2 vế trên lại ta được: \(\left(a-1\right)^3+\left(b-2\right)^3+5a+5b-15=0\)
\(\Leftrightarrow\left(a-1+b-2\right)\left[\left(a-1\right)^2-\left(a-1\right)\left(b-2\right)+\left(b-2\right)^2\right]+5\left(a+b-3\right)=0\)
\(\Leftrightarrow\left(a+b-3\right)\left[\left(a-1\right)^2-\left(a-1\right)\left(b-2\right)+\left(b-2\right)^2+5\right]=0\)
Mà \(\left(a-1\right)^2-\left(a-1\right)\left(b-2\right)+\left(b-2\right)^2+5\)
\(=\left[\left(a-1\right)^2-\left(a-1\right)\left(b-2\right)+\frac{1}{4}\left(b-2\right)^2\right]+\frac{3}{4}\left(b-2\right)^2+5\)
\(=\left[a-1-\frac{1}{2}\left(b-2\right)\right]^2+\frac{3}{4}\left(b-2\right)^2+5>0\left(\forall a,b\right)\)
\(\Rightarrow a+b-3=0\Leftrightarrow a+b=3\)
Vậy a + b = 3