Question

factor x^n-a^n

Answer 1

\[x^n-a^n=(x-a)(x^{n-1}+x^{n-2}a+...xa^{n-2}+a^{n-1})\]

Answer 2

need help understanding how this works.

Answer 3

@ganeshie8

Answer 4

are you looking for a proof or just an intution of why it works ?

Answer 5

anything

Answer 6

\({

(x-a)(x^{n-1}+x^{n-2}a+...xa^{n-2}+a^{n-1})\\~\\
=x(x^{n-1}+x^{n-2}a+...xa^{n-2}+a^{n-1})\\
-a(x^{n-1}+x^{n-2}a+...xa^{n-2}+a^{n-1})\\~\\


\begin{align}= x^{n} &+ x^{n-1} a &+ x^{n-2} a^2 &+ &\cdots &&+ x a^{n-1} &
\\ &- x^{n-1} a &- x^{n-2} a^2 &-& \cdots&& - x a^{n-1} &- a^{n}

\end{align}}
\)

Answer 7

i have just expanded
add them vertically, everything cancels out except for the first and last terms

Answer 8

let me knw if you have any questions

Answer 9

by simply multiplying them out i get distrubting.

im getting.

\[x^n+x^{n-1}a-x^{n-1}a-x^{n-2}a^2+x^2a^{n-2}+xa^{n-1}-xa^{n-1}-a^n\]

=\[x^n-x^{n-2}a^2+x^2a^{n-2}-a^n\]

Answer 10

you have forgotten \(\cdots\)

Answer 11

\(\cdots \) represents all the middle terms

Answer 12

for example :
\[\large x^5-a^5 = (x-a)(x^4 + x^3a + x^2a^2 + xa^3+a^4)\]

Answer 13

another example :
\[\large x^6-a^6 = (x-a)(x^5 + x^4a + x^3a^2 + x^2a^3+xa^4+a^5)\]

Answer 14

try expanding out right side of both above examples
and verify whether you get the left hand side or not