Question

[fan FA]

Cho \(am^3=bn^3=cp^3\)và \(\frac{1}{m}+\frac{1}{n}+\frac{1}{p}=1\)

Chứng minh rằng \(\sqrt[3]{a}+\sqrt[3]{b}+\sqrt[3]{c}=\sqrt[3]{am^2+bn^2+cp^2}\).

Answer 1

[Nguyễn Thị Thùy Dương]

=>\(am^3=bn^3=cp^3=\frac{am^3}{m}+\frac{bn^3}{n}+\frac{cp^3}{p}\)

=>\(am^3=bn^3=cp^3=am^2+bn^2+cp^2\)

\(\sqrt[3]{am^2+bn^2+cp^2}=m\sqrt[3]{a}=n\sqrt[3]{b}=p\sqrt[3]{c}\)

=>\(\sqrt[3]{am^2+bn^2+cp^2}.1=m\sqrt[3]{a}.\left(\frac{1}{m}+\frac{1}{n}+\frac{1}{p}\right)=\frac{m\sqrt[3]{a}}{m}+\frac{n\sqrt[3]{b}}{n}+\frac{p\sqrt[3]{c}}{p}\)

\(\sqrt[3]{am^2+bn^2+cp^2}=\sqrt[3]{a}+\sqrt[3]{b}+\sqrt[3]{c}\)