Gọi 3 số đó là a,b,c thì a + b + c = 0
+ \(a^3+b^3+c^3-3abc\)
\(=\left(a+b\right)^3+c^3-3ab\left(a+b\right)-3abc\)
\(=\left(a+b+c\right)\left[\left(a+b\right)^2-\left(a+b\right)c+c^2\right]-3ab\left(a+b+c\right)\)
\(=\left(a+b+c\right)\left(a^2+2ab+b^2-ac-bc+c^2-3ab\right)\)
\(=0\)
\(\Rightarrow a^3+b^3+c^3=3abc\)
\(=3\cdot48=144\)