p: "If x is a real number such that x3+4x=0, then x is 0′′ Let q: x is a real number such that x3+4x=0 r: x is 0(i) To show that statement p is true, we assume that q is true and then show that r is true. Therefore, let statement q be true.∴x3+4x=0 x(x2+4)=0 ⇒x=0 or x2+4=0However, since x is real, it is 0.Thus, statement r is true.Therefore, the given statement is true.(ii) To show statement p to be true by contradiction, we assume that $p$ is not true.Let x be a real number such that x3+4x=0 and let x is not 0.Therefore, x3+4x=0 x(x2+4)=0 x=0 or x2+4=0 x=0 or x2=−4 However, x is real. Therefore, x = 0, which is a contradiction since we have assumed that x is not 0Thus, the given statement p is true.(iii) To prove statement p to be true by contrapositive method, we assume that r is false and prove that q must be false.Here, r is false implies that it is required to consider the negation of statement r. This obtains the following statement.\\sim r: x is not 0It can be seen that (x2+4) will always be positive.x=0 implies that the product of any positive real number with x is not zero.Let us consider the product of x with (x2+4).∴x(x2+4)=0 ⇒x3+4x=0 This shows that statement q is not true.Thus, it has been proved that∼r⇒∼q Therefore, the given statement p is true.