\(\left(3x+1\right)^8>=0\)
\(\left(2y^2-32\right)^8>=0\)
Do đó: \(\left(3x+1\right)^8+\left(2y^2-32\right)^8>=0\)
Dấu '=' xảy ra khi 3x+1=0 và 2y2-32=0
\(\Leftrightarrow\left\{{}\begin{matrix}x=-\dfrac{1}{3}\\y\in\left\{4;-4\right\}\end{matrix}\right.\)