Compare Quadratic Expressions: a(a-b)² vs a(a+b)²-ab²

The task is to determine the inequality $a(a-b)^2$ compared to $a(a+b)^2 - ab^2$ given $0 < a, b$. Here's how to solve it:

First, let's expand and simplify each expression:

Starting with $a(a-b)^2$, expand as follows:

• $(a-b)^2 = a^2 - 2ab + b^2$.
• Multiplying by $a$ gives: $a(a^2 - 2ab + b^2) = a^3 - 2a^2b + ab^2$.

Next, expand $a(a+b)^2-ab^2$:

• $(a+b)^2 = a^2 + 2ab + b^2$.
• Multiplying by $a$ gives: $a(a^2 + 2ab + b^2) = a^3 + 2a^2b + ab^2$.
• Subtracting $ab^2$ yields: $a^3 + 2a^2b + ab^2 - ab^2 = a^3 + 2a^2b$.

Now compare the two expressions:

• $a(a-b)^2 = a^3 - 2a^2b + ab^2$.
• $a(a+b)^2 - ab^2 = a^3 + 2a^2b$.
• The inequality $a(a-b)^2 ? a(a+b)^2 - ab^2$ is simplified to:\)
• $a^3 - 2a^2b + ab^2 < a^3 + 2a^2b$.
• Cancel $a^3$ from both sides: $-2a^2b + ab^2 < 2a^2b$.
• Bring like terms together: $-4a^2b + ab^2 < 0$.
• Factor $b$: $b(-4a^2 + ab) < 0$.

Given $b > 0$, divide through by $b$:

$-4a^2 + ab < 0$, or equivalently $-4a^2 < -ab$, hence $4a > b$.

Therefore, $a > \frac{b}{4}$.

The inequality holds as $a(a-b)^2 < a(a+b)^2 - ab^2$ when $a > \frac{b}{4}$.

Therefore, the correct choice is:

$<$ When $a > \frac{b}{4}$