The given compound statement is of the form “if p then q” p: x ∈ R such that $x^3$ + 4x = 0 q: x = 0 (i) Direct method: We assume that p is true, then x ∈ R such that $x^3$ + 4x = 0 ⇒ x ∈ R such that x($x^2$ + 4) = 0 ⇒ x ∈ R such that x = 0 or $x^2$ + 4 = 0 ⇒ x = 0 ⇒ q is true. So, when p is true, q is true. Thus, the given compound statement is true. (ii) Method of contradiction : We assume that p is true and q is false, then x ∈ R such that $x^3$ + 4x = 0 ⇒ x ∈ R such that x($x^2$ + 4) = 0 ⇒ x ∈ R such that x = 0 or $x^2$ + 4 = 0 ⇒ x = 0. which is a contradiction. So, our assumption that x ≠ 0 is false. Thus, the given compound statement is true. (iii) Method of contrapositive: We assume that q is false, then x ≠ 0 x ∈ R such that $x^3$ + 4x = 0 ⇒ x ∈ R such that x = 0 or $x^2$ + 4 = 0 ∴ statement q is false, so x ≠ 0. So, we have, x ∈ R such that x2 = –2 Which is not true for any x ∈ R. ⇒ p is false So, when q is false, p is false. Thus, the given compound statement is true.