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?

  • $\begingroup$ Please clear up your question. Even without using Latex, it is completely unclear what option (4) or option 2 are, what the premise of the problem is, etc. For your equations, all you need for Latex is start a new line and wrap your math with $$ math here $$. $\endgroup$
    – Steve Heim
    Nov 1, 2018 at 7:53
  • $\begingroup$ @SteveHeim I am referring to the options in the question. $\endgroup$ Nov 1, 2018 at 8:06
  • 1
    $\begingroup$ Ah I see it now. You should transcribe your question to math.SE instead of putting a link, which will eventually be inaccessible. It also makes your question much more readable. $\endgroup$
    – Steve Heim
    Nov 1, 2018 at 8:06
  • $\begingroup$ I edited the latex of your question. You should be able to edit your own answer to see how it looks once it has been applied. $\endgroup$
    – Steve Heim
    Nov 1, 2018 at 8:23
  • 2
    $\begingroup$ Why does $ab=6x^2$ split in the way you describe? There are lots of products that give you $6x^2$, for instance $1/2$ times $12x^2$, $-2x$ times $-3x$ (and infinitely more options). This is where your error arises, try, instead to substitute for $b$ at this point. $\endgroup$ Nov 1, 2018 at 8:24

3 Answers 3


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.

  • $\begingroup$ Thanks!But,where am I doing error? $\endgroup$ Nov 1, 2018 at 8:16
  • $\begingroup$ I’ve edited the answer to include the reason your answer isn’t correct. Michael Burr’s comment sums it up nicely: it’s not possible to just “determine” what $a$ and $b$ can or cannot be. You must carry out substitution to get the answer satisfying all given conditions. $\endgroup$
    – KM101
    Nov 1, 2018 at 8:38

You're losing signs in your division. You actually also have the options

$$ ab = (-3x)(-2x)= 6x^2 \text{ or } ab = (-2x)(-3x) = 6x^2 $$

To check, you need to plug all of these into your original constraint. This is easy to do since you know that $ b = a -x$. With $a = 2x$ you get $b=-x$, and if you plug this into your second constraint you get

$$ (2x)^3 - (-x)^3 = 8x^3 + x^3 \neq 19x^3 $$

I'll leave it to you to test $a=-2x$.

  • $\begingroup$ Along with this sign case,I also missed few other cases like (3)(2x^2) or (2)(3x^2). Am I correct? $\endgroup$ Nov 1, 2018 at 8:14
  • $\begingroup$ @NavneetKumar Specifically, you can discard any cases where a or b are not proportional to x, as you can’t get both a single linear term by subtraction and a single quadratic term by multiplication in any other way. $\endgroup$
    – alex_d
    Nov 1, 2018 at 18:09

The work you did till you reached ab=6x^2 is all good But thats when you went wrong you chose a=3x and a=2x For a=3x b=2x its all good but for a=2x b=a-x=x so ab=2x^2 in this case you are wrong so what solves this is a=-2x In general for you to be able to solve such equation without guessing you just substitute b by its value: a(a-x)=6x^2 a^2-ax=6x^2 a^2-ax-6x^2=0 You want the value of a so you solve this equation for a: Delta=b^2-4ac=x^2+24x^2=25x^2 So a=3x or a=-2x


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