
We can solve the by the method of contradiction. Let us assume that 3✓25✓7 is a rational number. A rational number is always in the form of p/q where p and q are intergers, p and q are co prime and q not equal to zero. Therefore, 3✓25✓7 = p/q Squaring both sides (3✓25✓7)^2 =(p/q)^2 193  30✓14= p^2/q^2 30✓14= p^2/q^2 193 30✓14=p^2193q^2/q^2 ✓14=p^2193q^2/30q^2 Notice that LHS is irrational(14 is not a perfect square) but RHS is rational. Hence this situation is not possible. Hence proved.