# Factor $(y^{3} - 125)$

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?

• $a^3 - b^3 = (a-b)(a^2 +ab +b^2)$ and in general $$a^n -b^n = (a-b)(a^{n-1} +a^{n-2}b + a^{n-3}b^2 + \cdots +a^2b^{n-3} +ab^{n-2} +b^{n-1}) =(a-b) \sum_{k=1}^n a^{n-k}b^{k-1}$$ – azif00 Jul 17 '19 at 1:54
• It’s the difference of two cubes – J. W. Tanner Jul 17 '19 at 2:11

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.

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

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

• I see, and 125 is just $y^{3}$ – Burt Jul 17 '19 at 1:57

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:

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

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

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:

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