Prove that $a^3

Question

I want to prove, for all $a,b\in\mathbb{R}$, that $a^3<b^3$ if and only if $a<b$.

Assuming $a<b$, I need to prove that $b^3-a^3>0$.

I know that $$ b^3-a^3=(b-a)(a^2+b^2+ab) $$ and that $(b-a)>0$. However, I'm not sure how to conclude that $(a^2+b^2+ab)>0$. It's the $ab$ term causing the problem.

Can anyone provide a hint?

Do you know how to prove that $x^2 + x + 1 >0 $ for all real $x$? — Ben Martin, Apr 17 '22 at 07:55
Try breaking it into cases. Suppose $0\le a<b$. Then $ab\ge 0$. Suppose $a<0<b$. Then $a^3<0<b^3$. Suppose $a<b\le 0$. Then $ab\ge 0$. — SlipEternal, Apr 17 '22 at 07:55

score 0 · Answer 1 · answered Apr 17 '22 at 08:02

Assume $a \neq 0$. Then yoi can write $(a^2+ab+b^2)$ as $a^2(t^2+t+1)$ where $t=\frac{b}{a}$. Since the $t^2+t+1>0$ for every $t\in \mathbb R$ the conclusion follows. On the other side, if $a=0$ then you get $b^2>0$ wich is true unless $b=0$. But in this case you have also that $a=b$.

Prove that $a^3

1 Answers1