Editing Difference of two squares (section)

==Generalizations==
[[Image:Rhombus understood analytically.svg|thumb|right|Vectors {{math|'''a'''}}&nbsp;(purple), {{math|'''b'''}}&nbsp;(cyan) and {{math|'''a''' + '''b'''}}&nbsp;(blue) are shown with [[arrow (symbol)|arrows]]]]
The identity also holds in [[inner product space]]s over the [[field (mathematics)|field]] of [[real numbers]], such as for [[dot product]] of [[Euclidean vector]]s:
:<math>{\mathbf a}\cdot{\mathbf a} - {\mathbf b}\cdot{\mathbf b} = ({\mathbf a}+{\mathbf b})\cdot({\mathbf a}-{\mathbf b})</math>
The proof is identical. For the special case that {{math|'''a'''}} and {{math|'''b'''}} have equal [[normed vector space|norms]] (which means that their dot squares are equal), this demonstrates [[analytic geometry|analytically]] the fact that two diagonals of a [[rhombus]] are [[right angle|perpendicular]].  This follows from the left side of the equation being equal to zero, requiring the right side to equal zero as well, and so the vector sum of {{math|'''a''' + '''b'''}} (the long diagonal of the rhombus) dotted with the vector difference {{math|'''a''' − '''b'''}} (the short diagonal of the rhombus) must equal zero, which indicates the diagonals are perpendicular.

===Difference of two ''n''th powers===
[[File:Difference_of_squares_and_cubes_visual_proof.svg|thumb|Visual proof of the differences between two squares and two cubes]]
If {{mvar|a}} and {{mvar|b}} are two elements of a commutative ring, then<ref>{{cite book |last=Grigorieva |first=Ellina |title=Methods of Solving Nonstandard Problems |isbn=978-3-319-19886-6 |doi=10.1007/978-3-319-19887-3_2 |publisher=Birkhäuser |at=Eq. 2.13, {{pgs|83}} |year=2015 }}</ref>

<math display=block>a^n-b^n=(a-b)\biggl(\sum_{k=0}^{n-1} a^{n-1-k}b^k\biggr).</math>

The second factor looks similar to the [[binomial theorem|binomial expansion]] of <math>(a+b)^{n-1}</math>, except that it does not include the [[binomial coefficient]]s {{tmath|\tbinom{n-1}{k} }}.