They Must All Be True Right?

Relevant wiki: Factorization of Polynomials

Case 1 : Suppose the first equation ( $x=y$ ) is true, then squaring and cubing this equation shows that the other two equations are true as well.

Case 2 : Suppose the second equation ( $x^2=y^2$ ) is true, then by difference of two squares , we have $x^2 - y^2= 0 \Rightarrow (x-y) (x+y) = 0 \Rightarrow x= \pm y$ . By zero product property (ZPP) , we have $x =y$ or $x=-y$ .

Subcase 2.1 : If $x=y$ , then we're back to Case 1. And so it is possible to have all the given equations to be true.

Subcase 2.2 : If $x=-y$ , then the first equation is not true. Similarly, by cubing this equation, we have $x^3 = -y^3$ , so the third equation is not true as well. Hence, it is possible that the second equation is true only.

Case 3 : Suppose the third equation ( $x^3=y^3$ ) is true, then applying the perfect cube identity gives $x^3 - y^3 = 0\Rightarrow (x-y)(x^2 + xy + y^2) = 0$ . By (ZPP), we have $x - y = 0$ or $x^2 + xy + y^2= 0$ .

Subcase 3.1 : If $x= y$ , then we're back to Case 1. And so it is possible to have all the given equations to be true.

Subcase 3.3 : If $x^2 + xy + y^2 = 0$ , then by solving this quadratic equation in terms of $x$ , we get a quadratic discriminant of $(-y)^2 - 4(y^2) = -3y^2 \leq 0$ . Since $x$ and $y$ are real numbers, then $y = 0$ only, this leaves us with $x = 0$ as well. Hence, it is possible to have all the given equations to be true.

In conclusion, by comparing all these cases, all of these cases share a common fact: Only The second equation is definitely true.

CASE-1 If x=y....then on squaring both sides...x^2=y^2 and on taking cube on both sides....x^3=y^3...both the obtained equations are true

CASE-2 If x=-y...... then on squaring both sides......x^2=y^2 and on taking cube on both.........x^3= -y^3......which is not true.

x² = y² : positive and negative numbers which are equally far from zero have equal squares

If X = -y, then x^2 = y^2 but x^3 does not equal y^3