Let a and b be positive real numbers, m = a^3 + b^3 and n = a^2b + ab^2. Compare m and n.

Step-by-step solution

1. 1

   Compare by subtracting, not by dividing. For two expressions the reliable move is to study the sign of $m-n$: if $m-n>0$ then $m>n$, if $m-n=0$ they are equal, and if $m-n<0$ then $m<n$. Division would also work here since $a,b>0$, but subtraction keeps everything polynomial and factorable.

2. 2

   Form the difference and group in pairs.

   $m-n=a^3+b^3-a^2b-ab^2$

   Group the cubes with the mixed terms that share a variable:

   $m-n=(a^3-a^2b)+(b^3-ab^2)=a^2(a-b)+b^2(b-a)$

3. 3

   Pull out the common factor $(a-b)$. Since $b-a=-(a-b)$,

   $m-n=a^2(a-b)-b^2(a-b)=(a-b)(a^2-b^2)$

   and $a^2-b^2=(a-b)(a+b)$, so

   $\boxed{m-n=(a+b)(a-b)^2}$

   This factored form is the whole solution: it turns a comparison into a sign inspection.

4. 4

   Read off the sign. With $a>0$ and $b>0$ we have $a+b>0$, and $(a-b)^2\ge 0$ for any real numbers. A positive number times a nonnegative number is nonnegative, so

   $m-n\ge 0\qquad\Longrightarrow\qquad m\ge n$

5. 5

   Pin down when equality happens. The product $(a+b)(a-b)^2$ is zero only if one factor is zero. Since $a+b>0$, equality forces $(a-b)^2=0$, that is $a=b$. So

   $a^3+b^3>a^2b+ab^2\ \text{ when } a\neq b,\qquad a^3+b^3=a^2b+ab^2\ \text{ when } a=b$

6. 6

   Check with numbers. Take $a=3,b=1$: $m=27+1=28$ and $n=9+3=12$, and indeed $(a+b)(a-b)^2=4\cdot 4=16=28-12$. Take $a=b=2$: $m=16$ and $n=16$, equal as predicted. The same identity also proves the familiar rearrangement fact that $a^3+b^3\ge ab(a+b)$ for positive $a,b$.