# Homework Help: How to factorize(y + 3)^3 + 8

1. Mar 8, 2008

### alpha01

$$(y + 3)^3 + 8$$

my attempt has led me to

(y+3)(y+3)(y+3+8)

but i doubt this is correct

2. Mar 8, 2008

### tiny-tim

Hi alpha01!

If the answer isn't obvious, then simplify the question as much as possible, then try to solve it, then "un-simplify" it back.

Put z = y + 3.

Then the question is: factorise $$z^3 + 8$$.

Can you do that?

(Hint: when is it zero?)

3. Mar 8, 2008

### alpha01

expanding and simplify the (y+3)^3:

(y^2 + 6y + 9)(y + 3)

(y^2 + 6y + 9) (y + 3 + 8)?

4. Mar 8, 2008

### alpha01

Please post the solution and i will apply it to other examples

5. Mar 8, 2008

### tiny-tim

No no no … that is expanding but it's not simplifying!

"Expanding" means "longer", and "simplifying" means "shorter"!

$$z^3 + 8$$ is shorter and simpler than $$(y + 3)^3 + 8$$.

So: can you factorise $$z^3 + 8$$?

6. Mar 8, 2008

### alpha01

(z+2)(z+2)(z+2)

so the answer to the original question is:

(y+3+2)(y+3+2)(y+3+2)

(y+5)(y+5)(y+5)?

Last edited: Mar 8, 2008
7. Mar 8, 2008

### tiny-tim

No.

Hint: can you see a number for which $$z^3\,+\,8\,=\,0$$?

(In other words, a root of $$z^3\,+\,8\,=\,0$$)

Put "z - " in front of it, and that gives you a factor of $$z^3\,+\,8\,=\,0$$!

(if you still can't get it, I'll give you the answer)

8. Mar 8, 2008

no..

9. Mar 8, 2008

### Feldoh

Solve for z:
$$z^3\,+\,8\,=\,0$$

You need to think of this as

$$z^3+x^3 = 0$$

Where x^3 = 8

Last edited: Mar 8, 2008
10. Mar 8, 2008

### alpha01

z = -6

11. Mar 8, 2008

### Feldoh

How are you getting z = -6?

$$z^3\,+\,8\,=\,0$$

$$z^3\,=\,-8$$

$$z\,=\,(-8)^{1/3}$$

12. Mar 8, 2008

### kbaumen

To solve this one, you should know this formula: a^3 + b^3 = (a + b)(a - ab + b). I hope that helps.

13. Mar 8, 2008

### alpha01

thank you.. thats all i needed

14. Mar 8, 2008

### alpha01

i meant to write -2

i know how to solve this now anyway

15. Mar 8, 2008

### alpha01

So, putting (y+3)^3 +8 in the form (a+b)(a^2 -ab+b^2):

$$((y+3)+2) ((y+3)^2 -2(y+3)+4)$$

but i dont think this is completely factorized?

16. Mar 9, 2008

### Gib Z

I'm not trying to be rude, but from posts 1, 3 and 6 it seems to me you really need to go back and brush up on your algebra.

17. Mar 9, 2008

### tiny-tim

Well, you must then expand the two main brackets so that there's no brackets inside them!

At that point, you've finished - it can't be factorised any further.

[size=-2](Unless you know about complex numbers.)[/size]

Though, just to check, you should then multiply it out again to make sure you get back to the original!

Are you still worried about anything?

18. Mar 9, 2008

### alpha01

I'm not trying to be rude, but posts 1, 3, 6 are all direct attempts of factorization. My general algebraic skills are fine (19/20 of a algebra quiz (no factorization questions) supports this).

tiny-tim, thanks all solved.

Last edited: Mar 9, 2008
19. Mar 9, 2008

### alpha01

ok heres post #3

(y+3)^3 expands to:

(y^2 + 6y + 9)(y + 3)

...

So can you please explain (with detailed solution) how this is apparently incorrect (and yes i know it can be expanded further i deliberately didn't expand it further.)

and yes i know what "simplifying" is, i didnt simplify because i hadn't yet fully expanded it and that would take it out of the ()() form, which is obviously not going to factorize it.

Last edited: Mar 9, 2008
20. Mar 9, 2008

### HallsofIvy

It certainly isn't correct because you can't take the "8" inside one factor. That 8 would now be multiplied by the other "(y+ 3)" terms and it isn't in the original form.

What you should have seen immediately was that (y+ 3)3+ 8= (y+3)3+ 23. Do you know a general formula for factoring a3+ b3? If you are asked to do a problem like this, I suspect it is introduced in this section!

21. Mar 9, 2008