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 $$3x$$.

• 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 $$ math here $$. – Steve Heim Nov 1 '18 at 7:53
• @SteveHeim I am referring to the options in the question. – Navneet Kumar Nov 1 '18 at 8:06
• 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. – Steve Heim Nov 1 '18 at 8:06
• 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. – Steve Heim Nov 1 '18 at 8:23
• 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. – Michael Burr Nov 1 '18 at 8:24

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 $$a=-2x$$.

• Along with this sign case,I also missed few other cases like (3)(2x^2) or (2)(3x^2). Am I correct? – Navneet Kumar Nov 1 '18 at 8:14
• @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. – alex_d Nov 1 '18 at 18:09

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 $$0$$.

$$a = 3x \text{ or } a = -2x$$

$$ab = 6x^2$$

You can’t just jump to conclusions on what $$a$$ and $$b$$ can be. Here, make the substitution $$\color{blue}{b = a-x}$$.

$$a\color{blue}{(a-x)} = 6x^2$$

$$a^2-ax = 6x^2 \implies a^2-ax-6x^2 = 0$$

• 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. – KM101 Nov 1 '18 at 8:38