# How to factorize(y + 3)^3 + 8

• alpha01
The expanded form of (y+3)^3 + 8 is (y+3)(y+3)(y+3) + 8. When simplified, it becomes (y+3)3 + 8, which is (y+3)(y+3)(y+3 + 8). Keep simplifying and you will end up with (y+3+2)(y+3+2)(y+3+2). This is the most simplified form, which can also be written as (y+5)(y+5)(y+5). In summary, the expanded form of (y+3)^3 + 8 is (y+5)(y+5)(y+5), which can
alpha01
$$(y + 3)^3 + 8$$

my attempt has led me to

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

but i doubt this is correct

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?)

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

(y^2 + 6y + 9)(y + 3) so is the answer:

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

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

alpha01 said:
expanding and simplify the (y+3)^3:

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

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 $$z^3 + 8$$?

(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:
alpha01 said:
(z+2)(z+2)(z+2)

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)

no..

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:
z = -6

How are you getting z = -6?

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

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

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

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

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

thank you.. that's all i needed

Feldoh said:
How are you getting z = -6?

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

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

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

i meant to write -2

i know how to solve this now anyway

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 don't think this is completely factorized?

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.

alpha01 said:
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 don't think this is completely factorized?

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?

Gib Z said:
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.

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:
Gib Z said:
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.

ok here's 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:
alpha01 said:
$$(y + 3)^3 + 8$$

my attempt has led me to

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

but i doubt this is correct
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!

alpha01 said:
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 don't think this is completely factorized?

## 1. What does it mean to factorize (y + 3)^3 + 8?

Factorizing is the process of breaking down an expression into smaller parts that can be multiplied together to get the original expression. In this case, (y + 3)^3 + 8 can be factored into (y + 3)(y + 3)(y + 3) + 8.

## 2. What are the steps to factorize (y + 3)^3 + 8?

The first step is to identify if there are any common factors that can be pulled out of the expression. In this case, there are no common factors between (y + 3)^3 and 8. Next, we can use the formula (a + b)^3 = a^3 + 3a^2b + 3ab^2 + b^3 to expand (y + 3)^3. This gives us (y^3 + 9y^2 + 27y + 27) + 8. Finally, we can combine like terms to get the final factorized form of (y + 3)(y^2 + 9y + 35).

## 3. Can (y + 3)^3 + 8 be factored further?

No, (y + 3)^3 + 8 is already in its simplest factorized form. The expression cannot be broken down into smaller parts without changing its meaning.

## 4. What are the applications of factorizing expressions?

Factorizing is used in various fields such as mathematics, physics, and engineering. It helps in simplifying complex expressions, solving equations, and finding the roots of polynomials. It is also used in simplifying problems in chemistry and economics.

## 5. What are some tips for factoring expressions like (y + 3)^3 + 8?

One helpful tip is to always look for common factors first. If there are no common factors, try to use known formulas to expand the expression. It is also helpful to practice factoring different types of expressions to improve your skills.

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