Show tha

(i) Direct method: x³ + 4x = 0 or x(x² + 4) = 0
Now x² + 4 ≠ 0 and x ∈ R and hence x = 0.

(ii) Method of Contradiction: Let x ≠ 0 and let x = p, p ∈ R is a root of x³ + 4x = 0. Therefore p³ + 4p = 0
(or) p(p² + 4) = 0 as p = 0 Thus p² + 4 = 0 which is not possible. Therefore, our supposition is wrong. Hence p = 0 or x = 0
 
(iii) Contra positive method: p is not true

Let x = 0 is not true Let x = p ≠ 0

Therefore p³ + 4p = 0

P being the root of x² + 4 = 0

(or) p(p² + 4) = 0

Now p = 0, also (p² + 4) = 0

Which implies p(p² + 4) = 0

Now p = 0, also p² + 4 = 0

p(p² + 4) ≠ 0 if p is not true.

Hence x = 0 is the root of x³ + 4x = 0