Septic Factorization

Algebra Level 2

Which of these answer choices is the factorization of $(a+b)^7-a^7-b^7$ ?

7ab(a+b)(a^2+ab+b^2)^2

5(a+b)^2(a^2+ab+b^2)

5ab(a+b)^6

7ab(a+b)^5

3 solutions

Calvin Lin Staff
Aug 12, 2015

We will use the Remainder-Factor theorem to avoid expanding out all of the terms and trying to factorize them.

Consider $f(a,b) = (a+b)^7 - a^7 - b^7$ .

Since $f(0,b) = 0$ , this tells us that $a \mid f(a,b)$ .
Since $f(a,0) = 0$ , this tells us that $b \mid f(b,b)$ .
Since $f(a,-a) = 0$ , this tells us that $a+b \mid f(b,b)$ .
Since $f(a,\omega a) = 0$ , this tells us that $(a^2 + ab + b^2) \mid f(b,b)$ .

Thus, we have $f(a,b) = C ab(a+b)(a^2 + ab+b^2) \times g(a,b)$ , where $C$ is a constant and $g(a,b)$ is a degree 2 symmetric polynomial. The possibilities of $g(a,b)$ are $ab, (a+b)^2, (a^2 + ab+b^2)$ .

Now, consider $f'(a,b) = 7 ( a+b) ^ 6 - 7 b ^ 6$ . Once again, since $f' ( a, \omega a ) = 0$ , hence we can conclude that $(a^2 + ab + b^2 ) \mid f'(a,b)$ .

Thus, $f(a,b) = C ab(a+b)(a^2+ab+b^2) ^2$ .

Finally, setting $a = b = 1$ , we get that $C = 7$ .

Why and how did you choose the function $f'(a,b) = 7(a+b)^6 - 7b^6$ ?

Pi Han Goh - 5 years, 10 months ago

Because $ab, (a+b), (a^2 + ab+b^2)$ are already factors of $f(a,b)$ , and we have to determine the repeated factor.

To do so, we differentiate the function, in which case only repeated factors will still remain as factors.

E.g. we can check that $f'(0,b) \neq 0, f'(a, -a) \neq 0$ , and so we can conclude that $(ab)^2 \not \mid f(a,b)$ and $(a+b)^2 \not \mid f(a+b)$ .

Calvin Lin Staff - 5 years, 10 months ago

SIr when you differntiadied the above function shouldn't there be 7a^6 in the expression. P.S:-i am just a beginner in calculus

Rishi Sharma - 5 years, 10 months ago

@Rishi Sharma – To clarify: he is only differentiating with respect to $b$ . So $a^7$ vanishes after differentiation. It doesn't matter whether you differentiate with respect to $a$ or $b$ because it is a symmetric polynomial. That is, $f(a,b) = f(b,a)$ .

Pi Han Goh - 5 years, 10 months ago

@Pi Han Goh – Right. The "proper" way to apply Remainder-Factor Theorem, is actually to use it for the one-variable case (which is the only time that it is always true).

So, by right, the polynomial that we should consider is $f(x) = (1+x)^7 - 1 ^7 - x^7$ and factorize that accordingly.

[I just didn't want to go into that for the solution, to avoid further confusion.]

Calvin Lin Staff - 5 years, 10 months ago

Chew-Seong Cheong
Aug 11, 2015

$(a+b)^7-a^7-b^7 \\ = a^7 + 7a^6b + 21a^5b^2 + 35a^4b^3 + 35a^3b^4 + 21a^2b^5 + 7ab^6 + b^7 - a^7 - b^7 \\ = 7a^6b + 21a^5b^2 + 35a^4b^3 + 35a^3b^4 + 21a^2b^5 + 7ab^6 \\ = 7ab (a^5 +b^5) + 21a^2b^2(a^3 +b^3) + 35a^3b^3(a +b) \\ = 7ab [(a+b)(a^4 +b^4) - ab(a^3 +b^3)] + 21a^2b^2(a^3 +b^3) + 35a^3b^3(a +b) \\ = 7ab (a+b)(a^4 +b^4) + 14a^2b^2(a^3 +b^3) + 35a^3b^3(a +b) \\ = 7ab (a+b)(a^4 +b^4) + 14a^2b^2[(a+b)(a^2 +b^2) - ab(a+b)] + 35a^3b^3(a +b) \\ = 7ab (a+b)(a^4 +b^4) + 14a^2b^2(a+b)(a^2 +b^2) + 21a^3b^3(a +b) \\ = 7ab (a+b)[a^4 +b^4 + 2ab(a^2 +b^2) + 3a^2b^2] \\ = \boxed{7ab (a+b)(a^2 + ab + b^2)^2}$

Moderator note:

The remainder-factor theorem can be useful in this case to help us with this factorization.

Consider $f(x) = (1+x)^7 - 1^7 - x^7$ . Then, when $\omega$ is the complex cube root of unity, we have $f( \omega) = ( - \omega^2) ^7 - 1^7 - \omega^7 = -\omega^2 - 1 - \omega = 0$ . Hence, this tells us that $( x - \omega )$ divides $f(x)$ and thus $(x^2 + x + 1 ) \mid f(x)$ .

I also use this common factorization

Muh Amin - 5 years, 10 months ago

Jake Lai
Aug 9, 2015

Just check with $a = b = 1$ .

Moderator note:

How can we factorize this otherwise?

Yep. Who has the time to factorize? :P

Mehul Arora - 5 years, 10 months ago

I will note something quite nice: we have that the multiplicity of $p$ in the prime factorisation of $\binom{p}{k}$ is 1 exactly, which is why two of the options cannot be.

Jake Lai - 5 years, 10 months ago

Septic Factorization

3 solutions

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