Lời giải:
Đặt \(\sqrt[3]{4-\sqrt{15}}=m\)
Khi đó \(a=\frac{1}{m}+m\Rightarrow a^3-3a=\frac{1}{m^3}+\frac{3}{m}+3m+m^3-3(\frac{1}{m}+m)\)
\(=\frac{1}{m^3}+m^3=\frac{1}{4-\sqrt{15}}+4-\sqrt{15}=4+\sqrt{15}+4-\sqrt{15}=8(*)\)
Đặt \(\sqrt[3]{\frac{25+\sqrt{621}}{2}}=n; \sqrt[3]{\frac{25-\sqrt{621}}{2}}=p\)
\(\Rightarrow n^3+p^3=25; np=\sqrt[3]{\frac{25^2-621}{4}}=1\)
\(\Rightarrow (n+p)^3=n^3+p^3+3np(n+p)=25+3(n+p)\)
Do đó:
\(b^3-b^2=\frac{1}{27}(1-n-p)^3-\frac{1}{9}(1-n-p)^2\)
\(=\frac{1}{27}[1-3(n+p)+3(n+p)^2-(n+p)^3]-\frac{1}{9}[1-2(n+p)+(n+p)^2]\)
\(=\frac{-2}{27}+\frac{n+p}{9}-\frac{(n+p)^3}{27}\)
\(=\frac{-2}{27}+\frac{n+p}{9}-\frac{25+3(n+p)}{27}=-1(**)\)
Từ \((*);(**)\Rightarrow a^3+b^3-b^2-3a+100=8+(-1)+100=107\)
@Akai Haruma @Lightning Farron soyeon_Tiểubàng giải Nguyễn Huy Tú
