Elementary Level Algebra Question

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}

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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?

Answer 1

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.

Answer 2

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 $a=-2x$.

Answer 3

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