factorisation of cyclic expressions

[undone path]

need help factorising this

[versed canopy]

1. Wait patiently for a helper to come along.
2. Once someone helps you, say thank you and close the thread with:

+close

3. Feel free to nominate the person for helper of the week in #helper-nominations
4. Do not ping the mods, unless someone is breaking the rules.
5. If you're happy with the help you got here, and the server overall, you can contribute financially as well:

[weak pagoda]

We can note the following:
(a + b + c)^5 - a^5 - b^5 - c^5 = ((a + b + c)^5 - a^5) - (b^5 + c^5)
(a + b + c)^5 - a^5 - b^5 - c^5 = ((a + b + c)^5 - b^5) - (a^5 + c^5)
(a + b + c)^5 - a^5 - b^5 - c^5 = ((a + b + c)^5 - c^5) - (a^5 + b^5)
The first expression is clearly divisible by (b + c), the second by (a + c), the third by (a + b). But since this is all the same expression, that means the whole thing is divisible by (a + b)(a + c)(b + c).
So, we have:
(a + b + c)^5 - a^5 - b^5 - c^5 = (a + b)(a + c)(b + c)P(a, b, c)
Here P(a, b, c) is a homogeneous quadratic polynomial:
P(a, b, c) = Aa^2 + Bb^2 + Cc^2 + Uab + Vac + Wbc
But the initial expression is symmetric in a, b, c. So, we can simplify the polynomil a bit:
P(a, b, c) = A(a^2 + b^2 + c^2) + U(ab + ac + bc)
Thus, we get:
(a + b + c)^5 - a^5 - b^5 - c^5 = (a + b)(a + c)(b + c)(A(a^2 + b^2 + c^2) + U(ab + ac + bc))
There are many approaches to find A and U. I suggest first taking a = b = c = 1, then a = 0, b = c = 1. That will give you two linear equations in A and U.

Maybe not the best way to do this, but it should work.

[nova pilot]

This is exactly why i say your good at math 😑

[weak pagoda]

Ah, thanks 😅
Though, this would work in very specific cases, such as this one.

[young coyote]

@weak pagoda has given 1 rep to @echo lava

[undone path]

How did we take A and U common from those, 🙏🏻🙏🏻🙏🏻🙏🏻 thanks for the lovely answer brother but I am still kind of muddled at that step

[young coyote]

@undone path has given 1 rep to @weak pagoda

[weak pagoda]

As I said, you can make a system of two equations by substituting the following:

1. a = b = c = 1.
2. a = 0, b = c = 1.

[undone path]

I mean brother you have written above that P(a,b,c) = Aa² + Bb² + Cc² + Uab + Vac + Wbc

Aren't they all distinct coefficients?

How did we take A and U common from those

[weak pagoda]

Yes. But our initial polynomial is symmetric in a, b, c.

So, the coefficients of a^2, b^2 and c^2 must be the same. Same for ab, ac and bc.

[undone path]

Oh????!?(

[weak pagoda]

Thus, we get P(a, b, c) = A(a^2 + b^2 + c^2) + U(ab + ac + bc).

[undone path]

Thanks a lot 🙏🏻🙏🏻🙏🏻🙏🏻

[weak pagoda]

You're welcome!

[undone path]

Now I understood the assignment

🙏🏻🙏🏻🙏🏻

[nova pilot]

yeah @weak pagoda is good at math ( in my opinion )

thanks @weak pagoda

[young coyote]

@echo lava has given 1 rep to @weak pagoda