- #1

Hyperreality

- 202

- 0

Show that (x + a + b)^7 - x^7 - a^7 - b^7 is divisble by

x^2 + (a + b)x +ab.

You are using an out of date browser. It may not display this or other websites correctly.

You should upgrade or use an alternative browser.

You should upgrade or use an alternative browser.

- Thread starter Hyperreality
- Start date

- #1

Hyperreality

- 202

- 0

Show that (x + a + b)^7 - x^7 - a^7 - b^7 is divisble by

x^2 + (a + b)x +ab.

- #2

drag

Science Advisor

- 1,100

- 1

You can always solve the entire excercise...

Live long and prosper.

- #3

KLscilevothma

- 322

- 0

Show that (x + a + b)^7 - x^7 - a^7 - b^7 is divisble by x^2 + (a + b)x +ab.

Hint:

Notice that (x+a) and (x+b) are the 2 factors of x^2 + (a + b)x +ab.

So it is equivalent to show that (x + a + b)^7 - x^7 - a^7 - b^7 is divisible by both (x+a) and (x+b).

Let f(x) = (x + a + b)^7 - x^7 - a^7 - b^7

...

...

...

...

Can you continue from here?

Hope this help.

- #4

ottjes

- 24

- 0

eg. (x+a)^2=x^2+2xa+a^2

maybe rewrite some terms then and you will see that it is divisible by x^2 + (a + b)x +ab

- #5

Hurkyl

Staff Emeritus

Science Advisor

Gold Member

- 14,967

- 19

Writing it out however... *shudder* I wouldn't wish writing out a trinomial to the 7th power to anyone!

Hurkyl

Share:

- Last Post

- Replies
- 5

- Views
- 425

- Last Post

- Replies
- 3

- Views
- 543

- Replies
- 2

- Views
- 522

- Last Post

- Replies
- 5

- Views
- 629

- Last Post

- Replies
- 1

- Views
- 390

MHB
Squeeze Theorem

- Last Post

- Replies
- 3

- Views
- 550

- Replies
- 38

- Views
- 444

- Last Post

- Replies
- 31

- Views
- 2K

- Replies
- 1

- Views
- 217

- Replies
- 1

- Views
- 254