<s>[INST] Elementary Level Algebra Question I'm trying to solve the following homework problem:
If $a\neq b$, $a^3-b^3 = 19x^3$ and $a-b=x$, which of the following conclusions is correct?
\begin{align}
\text{(1) }& a = 3x \\
\text{(2) }& a = 3x \text{ or } a = -2x \\
\text{(3) }& a = -3x \text{ or } a = 2x \\
\text{(4) }& a = 3x \text{ or } a = 2x
\end{align}

I am getting option (4): $a = 3x$ or $a=2x$ as the answer, which is incorrect. The correct answer is $a=3x$ or  $a=-2x.$
My work:-
$$ (a-b)^3 + 3ab(a-b)=19x^3$$
$$x^3 + 3ab(x) = 19x^3$$
$$ab=6x^2$$
Hence by comparing we have,
$$a * b = 3x * 2x \text{  or  } a * b = 2x * 3x$$
So a can be either $2x$ or $3x$.
Why is my answer wrong?
 [/INST] Use difference of cubes.
$$a^3-b^3 = (a-b)(a^2+ab+b^2)$$
$$\implies (a-b)(a^2+ab+b^2) = 19x^3$$
Set $\color{purple}{a-b = x}$.
$$\implies \color{purple}{x}(a^2+ab+b^2) = 19x^3 \implies a^2+ab+b^2 = 19x^2$$
Set $\color{blue}{b = a-x}$.
$$\implies a^2+a\color{blue}{(a-x)}+\color{blue}{(a-x)}^2 = 19x^2$$
Move $19x^2$ to the LHS, expand, and simplify.
$$\implies a^2+a^2-ax+a^2-2ax+x^2-19x^2 = 0$$
$$\implies 3a^2-3ax-18x^2 = 0$$
$$\implies a^2-ax-6x^2 = 0$$
Factor the trinomial.
$$\implies (a-3x)(a+2x) = 0$$
Set either factor equal to $0$.
$$a = 3x \text{ or } a = -2x$$
Edit: You’ve asked where your error is. Your comparison part wasn’t correct.
$$ab = 6x^2$$
You can’t just jump to conclusions on what $a$ and $b$ can be. Here, make the substitution $\color{blue}{b = a-x}$.
$$a\color{blue}{(a-x)} = 6x^2$$
$$a^2-ax = 6x^2 \implies a^2-ax-6x^2 = 0$$
This leads to the same answer.
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