Why isn't $\frac{x^3+1}{x+1}$ equal to $x^2$?

Question

Can someone please tell me why

$$ \frac{x^3+1}{x+1} \neq x^2\;? $$

I thought you could expand the $x^3$ out like so and cancel it out, but I got it wrong on a test and I have not the slightest clue why.

$$ \frac{x\cdot x\cdot x+1}{x+1} $$

I just canceled out the $x+1$ from both the top and the bottom and got $x^2$ but this is wrong somehow?

Thanks for your help.

Order of operations means that you compute $x \cdot x \cdot x$ before adding $1$. It is different from $x \cdot x \cdot (x + 1) = x^2 (x + 1) = x^3 + x^2$. — , Sep 05 '17 at 22:15
$ \require{cancel}\dfrac{16}{64} = \dfrac{1\cancel{6}}{\cancel{6}4} = \dfrac{1}{4},$ right? — Xander Henderson, Sep 05 '17 at 22:49
I mean, you can always plug in a few numbers to see:$$\frac{0^3+1}{0+1}\ne0$$ — Simply Beautiful Art, Sep 05 '17 at 22:52

score 9 · Answer 1 · answered Sep 05 '17 at 22:37

I’ve said this before: if I were teaching at an elementary level, I would forbid the use of the word “cancel” in my classes. In dealing with fractions, the appropriate rule is that you may multiply (or divide) both top and bottom by the same nonzero quantity without changing the value of the fraction. And to a certain degree, this is the Cardinal Rule of Fractions. It’s what you do when you reduce, and it’s what you do when you try to add two fractions with different bottoms (“denominators”).

So, for instance, $x^3/2x=x^2/2$, by dividing top and bottom by $x$, but in the instant case, you have not divided top and bottom by anything to go from $x^3+1$ to $x+1$ upstairs, and (presumably) $x+1$ to $1$ downstairs.

Banish cancel from your vocabulary!

I really like this answer. It's good advice. (+1) – Sep 05 '17 at 22:57 — , Sep 05 '17 at 22:57

score 3 · Answer 2 · answered Sep 05 '17 at 22:18

Order of operations means that you compute $x \cdot x \cdot x$ before adding $1$. It is different from $x \cdot x \cdot (x + 1) = x^2 (x + 1) = x^3 + x^2$.

For an explicit answer, take $x = 0$ and notice that $$\frac{x^3 + 1}{x + 1} = \frac{0^3 + 1}{0 + 1} = \frac{0 + 1}{1} = \frac 1 1 = 1 \ne 0 = 0^2 = x^2.$$ I've included this many many steps so that you can see exactly which operation happens in what order.

score 1 · Answer 3 · answered Sep 05 '17 at 22:34

When you cancel in a fraction, essentially you're multiplying by $1$. Consider the fraction $$\dfrac {abc}{dag}$$ where $a,b,c,d,g$ are all just numbers. We then have $$\dfrac {abc}{dag}=\dfrac {abc}{dag}\cdot 1 =\dfrac {abc}{dag} \cdot \dfrac {\frac {1}{a}}{\frac {1}{a}} = \dfrac {\left ( \frac {abc}{a} \right)}{\left( \frac {dag}{a} \right)} = \frac {bc}{dg}$$

Try taking this approach to your problem and see what happens.

score 1 · Answer 4 · answered Sep 05 '17 at 23:28

Besides the comments about order of operations, always consider the following: Are the two equations equal for all $x\in\mathbb{R}$?

In this case an easy counter-example when $\frac{x^{3} + 1}{x+1} = x^{2}$ fails is when $x=-1$, then the LHS is undefined due to division by zero whereas the RHS is solved easily $x^2 = (-1)^{2} = 1$.

Emilija S. · Answer 5 · 2017-10-09T23:04:38.483

1

You can write $\frac{x^3 +1}{x+1}$ in the form: $\frac{x^3 +1}{x+1}=\frac{x^3 +1^3}{x+1}=$ $\frac{(x+1)\times (x^2-x+1)}{x+1}=$ $x^2-x+1$
which can be equal to $x^2$ only if $1-x$ is equal to $0$. It's possible only if $x=1$. And we have $x^2=1^2=1$ which is possible.

edited Oct 09 '17 at 23:04

answered Oct 08 '17 at 16:15

Emilija S.

111

This is incorrect. $a^3+b^3=(a+b)(a^2-ab+b^2)$ – Simply Beautiful Art Oct 09 '17 at 15:09
@Simply Beautiful Art Yeah I already fixed it. I think it's the real one now, isn't it? – Emilija S. Oct 09 '17 at 23:11
I must find it very strange that this was not mentioned before :-/ – Simply Beautiful Art Oct 09 '17 at 23:13
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Need to be more carefully i know, thank you for this time! – Emilija S. Oct 09 '17 at 23:34

Why isn't $\frac{x^3+1}{x+1}$ equal to $x^2$?

5 Answers5