Can anyone tell me how they expanded $x^n-a^n$?

Question

I am not getting how they factorised the $x^n-a^n$ term?

Answer 1

Lets do it backwards for a special case . You have the sum of powers

$$1+x+x^2+\cdots +x^{n-1}.$$

Multiply this by $x$ and you get

$$x+x^2+x^3+\cdots+x^n.$$

You find that increasing the exponent of each term by $1$ essentially did nothing else but removing the $1$ from the beginning and adding the $x^n$ at the end. Every term in between is contained in both sums. This means we can get the second form from the first one by the following operation instead:

$$x\cdot (1+x+\dots+x^{n-1})=(1+x+\dots+x^{n-1})+(x^n-1).$$

The left side is what we already did before, the right side expresses the removing of $1$ and adding of $x^n$ instead. Now do some rearrangement:

\begin{align} (1+x+\dots+x^{n-1})+(x^n-1) &= x\cdot (1+x+\dots+x^{n-1})\\ x^n-1 &= x\cdot (1+x+\dots+x^{n-1}) - (1+x+\dots+x^{n-1})\\ x^n-1 &= (x - 1)\cdot (1+x+\dots+x^{n-1}).\\ \end{align}

This already looks like the identity you are looking for, with the special assumption $a=1$. Lets generalize:

\begin{align} x^n-a^n &= a^n\left(\left(\frac xa\right)^n-1\right)\\ &\overset{(*)}= a^n\left(\frac xa-1\right)\left(1+\frac xa+\left(\frac xa\right)^2+\cdots +\left(\frac xa\right)^{n-1}\right) \\ &= a\left(\frac xa-1\right) \cdot a^{n-1}\left(1+\frac xa+\left(\frac xa\right)^2+\cdots +\left(\frac xa\right)^{n-1}\right) \\ &=(x-a)(a^{n-1}+ x a^{n-2} +\cdots x^{n-2}a + x^{n-1}). \end{align}

In step $(*)$ I used the special case and $x\mapsto \frac xa$ to expand the latter term.

Answer 2

Perhaps you are asking "How does someone come up with this factorization of $a^n-b^n$"?

If so, I would say the answer is: observation of patterns, guessing at generalizations, and proof. This is what mathematicians do.

Start with what we all know and maybe even discover for ourselves: $$a^2 - b^2 = (a-b)(a+b)$$ Next, take a harder one, but one which is still not too hard to know and perhaps discover: $$a^3 - b^3 = (a-b)(a^2+ab+b^2) $$

Next, try to observe a pattern. The first factor is $a-b$ for both of these. Then guess at a generalization: could there factorizations of similar form $$a^4 - b^4 = (a-b)(\text{some factor?}) $$ $$a^5 - b^5 = (a-b)(\text{some different factor?}) $$ and in general $$a^n - b^n = (a-b)(\text{some still different factor, but fitting some general pattern?}) $$ That's a very nice guess, but it's not yet completely mathematical until we can figure out a pattern for the second factor.

There are a couple of ways to proceed from here. One is to try to guess at a pattern for the second factor, and see if one can prove that this pattern is true in general. Each term of the second factor has degree $2$, the coefficient is $1$, and the pattern of exponents is to decrease the $a$-exponent one at a time and increase the $b$-exponent one at a time: $a^2 b^0 + a^1 b^1 + a^0b^2$. Could it be possible that the second factor of the $a^4 - b^4$ factorization is $a^3 b^0 + a^2 b^1 + a^1 b^2 + a^0 b^3$, or in short $$a^3 + a^2 b + a b^2 + b^3 $$ And then the completely general pattern would be $$a^{n-1} + a^{n-2} b + ... + a b^{n-2} + b^{n-1} $$ If so, perhaps it can be proved by induction. This is exactly what the other answers show.

Another way to proceed is to use the original guess that $a-b$ is the first factor, and then use long division of polynomials to find the second factor. That works too.

Answer 3

This is long division of polynomials. It essentially follows the same procedure of normal long division, where you divide the first term into the current term, distribute, and then subtract.

https://www.mathsisfun.com/algebra/polynomials-division-long.html

Since the answer was too short, I'll try to expand on it a bit.

We want to find $x-a\overline{)x^n - a^n}$

1. x divided into $x^n$ is $x^{n-1}$
2. $x^{n-1}$ is then distributed over (x-a): $x^{n-1}(x-a) = x^n-ax^{n-1}$
3. Subtract: $x^n-a^n-(x^n-ax^{n-1}) = ax^{n-1}-a^n$

Repeat.

$x-a\overline{)ax^{n-1}-a^n}$

1. x divided into $ax^{n-1}$ gives us $ax^{n-1}$
2. $ax^{n-1}$ is then distributed over (x-a): $ax^{n-1}(x-a)=ax^{n-1}-a^2x^{n-1}$
3. Subtract: $ax^{n-1}-a^n-(ax^{n-1}-a^2x^{n-1})=a^2x^{n-1}-a^n$

From this we can see that the power of a will increase by one each time, and the power of x will decrease by one, until we get to $a^n$.

So we have $x^{n-1}+ax^{n-2} + ... + xa^{n-2}+a^{n-1}$

Answer 4

$$x^{n}-a^{n}=\\x^{n}-a^{n}-a^{1}x^{n-1}+a^{1}x^{n-1}=\\ x^{n}-a^{1}x^{n-1}+a^{1}x^{n-1}-a^{n}=x^{n-1}(x-a)+a(x^{n-1}-a^{n-1})\\$$now do like this again $$x^{n-1}(x-a)+a(x^{n-1}-ax^{n-2}+ax^{n-2}-a^{n-1})\\ x^{n-1}(x-a)+a(x^{n-2}(x-a)+a(x^{n-2}-a^{n-2}))=\\ $$factor from $x-a$ $$(x-a)(x^{n-1}+x^{n-2}a^{1})+a^2(x^{n-2}-a^{n-2})=\\ (x-a)(x^{n-1}+x^{n-2}a^{1})+a^2(x^{n-2}-ax^{n-3}+ax^{n-3}-a^{n-2})=\\ (x-a)(x^{n-1}+x^{n-2}a^{1})+a^2(x^{n-3}(x-a)+a(x^{n-3}-a^{n-3}))$$ factor from $x-a$ $$(x-a)(x^{n-1}+x^{n-2}a^{1}+x^{n-3}a^{2})+a(x^{n-3}-a^{n-3})\\ (x-a)(x^{n-1}+x^{n-2}a^{1}+x^{n-3}a^)+a^3(x^{n-3}-ax^{n-4}+ax^{n-4}-a^{n-3})\\ (x-a)(x^{n-1}+x^{n-2}a^{1}+x^{n-3}a^{2}+x^{n-4}a^{3})+a^3(x^{n-3}-a^{n-3})\\$$ ans so on $$(x-a)(x^{n-1}+x^{n-2}a^{1}+x^{n-3}a^{2}+x^{n-4}a^{3}+x^{n-4}a^3+...+x^1a^{n-2}+...+a^{n-1})$$

Answer 5

The factorisation of $x^n-a^n$ is a high-school identity, derived from the factorisation $$1-u^n=(1-u)(1+u+u^2+\dots+u^{n-1})\tag{1}$$ which is proved by a simple induction (it is one of the best examples to make understand induction):

• First $(1)$ is true for $n=1$ or $2$.
• Second, suppose it is true for some $n\ge 1$. Then \begin{align} 1-u^{n+1}&=(1-u^n)+(u^n-u^{n+1})=(1-u)(1+u+u^2+\dots+u^{n-1})+u^n(1-u)\\&=(1-u)(1+u+u^2+\dots+u^{n-1}+u^n), \end{align} which ends the inductive step.

For the formula in case, set $u=\dfrac ax\enspace(\Leftrightarrow ux=a)$, and use $(1)$: \begin{align} x^n-a^n&=x^n(1-u^n)=x(1-u)x^{n-1}(1+u+u^2+\dots+u^{n-1})\\ &=(x-xu)(x^{n-1}+x^{n-2}xu+x^{n-3}x^2u^2+\dots+x^{n-1}u^{n-1}\\ &=(x-a)(x^{n-1}+x^{n-2}a+x^{n-3}a^2+\dots+a^{n-1}) \end{align}

Note that $(1)$, for $n=1$, can be written as the formula for the sum of the $n$ first terms of the geometric series: $$1+u+u^2+\dots+u^n=\frac{1-u^{n+1}}{1-u}.$$

Answer 6

Consider $S=x^{n-1}+ax^{n-2}+\dots+a^{n-2}x+a^{n-1}$. Then \begin{alignat}{30} xS&=x^n&{}+{}&ax^{n-1}&{}+{}&a^2x^{n-2}&{}+{}&\!\dotsb&{}+{}&a^{n-2}x^2&{}+{}&a^{n-1}x \\ aS&=&&ax^{n-1}&{}+{}&a^2x^{n-2}&{}+{}&\!\dotsb&{}+{}&a^{n-2}x^2&{}+{}&a^{n-1}x+a^n \end{alignat} Subtracting gives $$ xS-aS=x^n-a^n $$ so, for $x\ne a$, $$ S=\frac{x^n-a^n}{x-a} $$ When $n$ is a positive rational, do a change of variables. If $n=p/q$, set $x=y^q$ and $a=b^q$, so $$ \frac{x^n-a^n}{x-a}= \frac{y^p-b^p}{y^q-b^q}= \frac{y^{p-1}+by^{p-2}+\dots+yb^{p-2}+b^{p-1}} {y^{q-1}+by^{q-2}+\dots+yb^{q-2}+b^{q-1}} $$ after cancelling $y-b$ on top and bottom. For $y=b$, we get $$ \frac{pb^{p-1}}{qb^{q-1}}=\frac{p}{q}b^{(p-1)-(q-1)}=na^{n-1} $$ because $b=a^{1/q}$ and $$ \frac{(p-1)-(q-1)}{q}=\frac{p}{q}-1=n-1 $$