Arranging powers

• When x is negative, $x^2>x>x^3$ . For example, take $x=-2$ Then, $4>-2>-8$

• When x is positive and greater than 1, $x^3>x^2>x$ . For example when $x=2$ ,then $8>4>2$ . We can see any positive number greater than one always increases with greater powers.

• But the case $x>x^3>x^2$ is impossible.

Hence, only the first and third cases are possible.

Since $x,x^2,x^3$ are distinct real numbers so let's consider some of the following cases

case1 : If $x> 1$ , then it's always true that ${\color{#3D99F6}x<x^2<x^3}$ such as $2<4<8$ also if $x\in\mathbb Q$ or rational number then always true that $x> x^2> x^3$ or ${\color{#D61F06}x^3< x^2 <x}$ such as $\frac{1}{8}<\frac{1}{4} < \frac{1}{2}$ . (inequality reverses)

case 2 : if $x<-1$ then it's always true that $x^2 > x> x^3$ such as $9 > -3 > -27$ and also for negative rational numbers it always true that $x^2 > x^3 > x$

case 3: If $x >1$ is an irrational number then it true that ${\color{#3D99F6}x< x^2 < x^3}$ such as $\sqrt 2 < 2 <(\sqrt{2})^3$ and its also true that ${\color{#D61F06}x^3 < x^2 < x}$ such as $\frac{1}{(\sqrt 2)^3} < \frac{1}{2} < \frac{1}{\sqrt{2}}$

case 4: if $x< 1$ is irrational number then it will be similar to case 2:

Hence for distinct real numbers inequalities ${\color{#3D99F6}x < x^2 < x^3 }$ and ${\color{#D61F06} x^3 <x < x^2 }$ are always true .

I tried to graph the functions: f(x)=x, g(x)= $x^2$ , h(x)= $x^3$ , and then tried to find x values which satisfy the above inequalities. The first one is true for x>1 and the third one is true for x<-1. So only the 1st and 3rd answers are correct. I also found the: x, $x^2$ , $x^3$ inequality for -1<x<0 and the $x^3$ , $x^2$ , x, inequality for 0<x<1

• The first possibility, $x < x^2 < x^3$ will be true for any $x > 1$ , as higher powers of $x$ will get more and more positive.
• The third possibility, $x^3 < x < x^2$ will be true for any $x < \text{-}1$ ; the first inequality because $x^3$ will be more negative than $x$ , the second because $x^2$ will be positive.
• The second possibility, $x^2 < x^3 < x$ , can never be true; if we divide it by $x^2$ (as $x$ is clearly not zero), we get $1 < x < \frac{1}{x}$ , but if the first inequality is true, the second will definitely be false.

Firstly, $x = x^2 = x^3$ if $x = 0$ or $x = 1$ , so we must have $x \ne 0$ and $x \ne 1$ .

Inequality 1 $(x < x^2 < x^3)$ . Let $x = 2$ . Then $2 < 2^2 < 2^3$ and so the first inequality can be true.

Inequality 3 $(x^3 < x < x^2)$ . Let $x = -2$ . Then $(-2)^3 < -2 < (-2)^2$ and so the third inequality can be true.

Inequality 2 It remains to show whether or not $x^2 < x^3 < x$ for some $x \in \mathbb{R}$ .

Assume $a^2 < a^3 < a$ for some $a \in \mathbb{R}$ (and $a \ne 0$ , $a \ne 1$ ).

Since $a^2 < a^3 \Rightarrow a^2(a-1)>0$ , we have:

(1) $a^2 < 0$ and $a-1<0$

(2) $a^2>0$ and $a-1>0$ .

Because (1) is not possible, (2) implies $a \in (1, \infty)$ .

Since $a^3 < a \Rightarrow a(a^2-1)<0$ , we have:

(3) $a < 0 \cap a^2>1 \Rightarrow a \in (-\infty, -1)$ XOR

(4) $a > 0 \cap a^2<1 \Rightarrow a \in (0,1)$

Hence $a \in (-\infty, -1) \cup (0,1)$ . This is a contradiction, since $a^2 < a^3$ implied $a \in (1, \infty)$ ( $a$ cannot be a member of two disjoint sets).

There are great solutions already but I want to point out that you can divide through by x squared because it’s always positive so it won’t change the inequality but simplifies the problem. This simplifies all the inequalities and make them easier to examine.

You forgot the option $x^{3}$ < $x^{2}$ < x that can also be true (when x is a fraction between 0 and 1)

Idk if I did it right but I just broke it down to 4 ways x is positive and greater than 1 x^3>x^2>x x is positive and less than 1 x>x^2>x^3 x is negative and greater than -1 x^2>x^3>x^1 x is negative and less than -1 x^2>x>x^3 2 of these are there and the third one is not

x^3 < x^2 < x (If it's a fraction)