# Algebraic Step

1. Jan 19, 2005

Hello

I do not understand how to get from:

$$(x^2+1)^3 6x^2(x^3 - 1) + (x^3-1)^2 6x(x^2+1)^2$$

to $$6x(x^2+1)^2(x^3-1)(2x^3 + x - 1)$$

I tried factoring the $$6x$$ but have had no luck.

Any help is appreciated!

Thanks

Last edited: Jan 19, 2005
2. Jan 19, 2005

### MathStudent

try factoring 6x rather than 6x^2

3. Jan 19, 2005

### Hurkyl

Staff Emeritus
Well, you can't factor 6x^2, because the second term doesn't have an x^2!

Here, you're just trying to apply the distributive rule in reverse:

ab + ac = a(b + c)

What you need to do is identify the factors that appear in both terms. (e.g. x appears in both terms, but not x^2), and that's what you factor out. (that's the a).

4. Jan 19, 2005

### dextercioby

You can't factor $6x^{2}$,without getting something really ugly...

Pay attention with the calculations.The fact that u know the result already might help u if u don't see means of factorization...

Daniel.

5. Jan 19, 2005

### Curious3141

$$(x^2+1)^3 6x^2(x^3 - 1) + (x^3-1)^2 6x(x^2+1)^2 = (6x)(x^3 - 1)(x^2 + 1)^2\left( (x(x^2 + 1) + (x^3 - 1) \right) = 6x(x^2+1)^2(x^3-1)(2x^3 + x - 1)$$

Just group the terms and play with the algebra.

6. Jan 19, 2005

### Hurkyl

Staff Emeritus
Pfft, don't do the work for him -- one learns more when they do the work, rather than observing someone else's work.

7. Jan 19, 2005

### Curious3141

Point taken

8. Jan 19, 2005

### dextercioby

Not to mention u messed up the page layout... :grumpy:

Daniel.

9. Jan 19, 2005

(he does the work (singular) )

anyway i worked it out (like 1:20 AM here)

Thanks all

10. Jan 19, 2005

### Curious3141

Then set your browser differently. :tongue2: :rofl:

11. Jan 19, 2005

### futb0l

Up the resolution, it's just fine on 1280x1024 :P