Câu trả lời:

Ta có $a^3+b^3=(a+b)(a^2-ab+b^2)=a^2-ab+b^2$ ( vì a+b=1)

Lại có $2(a-b{)}^2\ge 0\iff 2a^2-4ab+2b^2\ge 0\iff 4a^2-4ab+4b^2\ge 2a^2+2b^2\iff 4(a^2-ab+b^2)\ge 2(a^2+b^2)\ge (a+b{)}^2=1\iff 4(a^2-ab+b^2)\ge 1\iff a^2-ab+b^2\ge \frac{1}{4}\Rightarrow a^3+b^3\ge \frac{1}{4}$

Vậy Min M=$\frac{1}{4}\iff a=b=\frac{1}{2}$