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I know this is a really elementary problem, but I can't seem to figure it out.
How do you factor: $(y^{3}-125)$?
The answer I got is $(y+5)(y^{2}-5y-25)$ but something about the signs just doesn't work.
How should this be factored?

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    $\begingroup$ $a^3 - b^3 = (a-b)(a^2 +ab +b^2)$ and in general $$a^n -b^n = (a-b)(a^{n-1} +a^{n-2}b + a^{n-3}b^2 + \cdots +a^2b^{n-3} +ab^{n-2} +b^{n-1}) =(a-b) \sum_{k=1}^n a^{n-k}b^{k-1}$$ $\endgroup$ – azif00 Jul 17 '19 at 1:54
  • $\begingroup$ It’s the difference of two cubes $\endgroup$ – J. W. Tanner Jul 17 '19 at 2:11
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This is just the difference of cubes.
The way to factor the difference of cubes is: $$(a^{3}-b^{3}) = (a-b)(a^{2}+ab+b^{2})$$.
In this case, your problem is really asking: $(y^{3}-5^{3})$. $y$ is $a$ and 5 is $b$. When you plug it all in, you get $(y-5)(y^{2}+5y+25)$. Your answer is almost right, but the signs are mixed up.

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You're almost there. It factors as $(y-5)(y^2 + 5y + 25)$.

In general, $x^n - y^n$ factors as $(x-y)(x^{n-1} + x^{n-2}y + \cdots + xy^{n-2} + y^{n-1})$.

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  • $\begingroup$ I see, and 125 is just $y^{3}$ $\endgroup$ – Burt Jul 17 '19 at 1:57
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In case you can't remember the formula for the difference of two cubes

(or that the complex cube roots $w$ of $1$ satisfy $w^2+w+1=0$),

you could note that $5$ is a zero of $y^3-125$, and that means $(y\color{red}-5)$ divides $y^3-125$.

Then you could do polynomial long division to get the other factor:

$(y^3-125)=y^2(y-5)+5y^2-125=y^2(y-5)+5y(y-5)+25y-125$

$=y^2(y-5)+5y(y-5)+25(y-5)=(y^2+5y+25)(y-5).$

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Pedestrian:

$P(y)=y^3-5^3$; $P(5)=0$, hence $(y-5)$ is a factor of the polynomial $P(y)$.

$P(y)=(y-5)Q(y)$, where $Q(y)$ is a polynomial of degree $2$.

$Q(y)=ay^2+by+c$;

$Q(y)=y^5-5^3=$

$(y-5)(ay^2+by +c);$

Comparing coefficients of the $2$ sides you get :

$a=1$; $c=5^2$;

No linear term on the left hand side:

$-5b+ c=0;$ $b= c/5= 5^2$.

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