Answer
This question can be proved by contradiction.
Let a>$\sqrt[3] n$ , b>$\sqrt[3] n$ , c>$\sqrt[3] n$ and a$\times$b$\times$c=n
Now a$\times$b$\times$c > n, this is a contradiction to the statement that a$\times$b$\times$c=n.
Therefore we can say that a$\leq$$\sqrt[3] n$ , b$\leq$$\sqrt[3] n$ , c$\leq$$\sqrt[3] n$.
Work Step by Step
This question can be proved by contradiction.
Let a>$\sqrt[3] n$ , b>$\sqrt[3] n$ , c>$\sqrt[3] n$ and a$\times$b$\times$c=n
Now a$\times$b$\times$c > n, this is a contradiction to the statement that a$\times$b$\times$c=n.
Therefore we can say that a$\leq$$\sqrt[3] n$ , b$\leq$$\sqrt[3] n$ , c$\leq$$\sqrt[3] n$.
