Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Difference between revisions of "Sum and difference of powers"

(have to go, open for improvisation.)
(→‎Differences of Powers)
 
(15 intermediate revisions by 9 users not shown)
Line 1: Line 1:
 
The '''sum and difference of powers''' are powerful factoring techniques that, respectively, factor a sum or a difference of certain powers.
 
The '''sum and difference of powers''' are powerful factoring techniques that, respectively, factor a sum or a difference of certain powers.
  
−
==Sums of Powers==
+
==Sums of Odd Powers==
 
<math>a^{2n+1}+b^{2n+1}=(a+b)(a^{2n}-a^{2n-1}b+a^{2n-2}b^2-\ldots-ab^{2n-1}+b^{2n})</math>
 
<math>a^{2n+1}+b^{2n+1}=(a+b)(a^{2n}-a^{2n-1}b+a^{2n-2}b^2-\ldots-ab^{2n-1}+b^{2n})</math>
  
 
==Differences of Powers==
 
==Differences of Powers==
−
If p is a positive integer and x and y are real numbers,
+
If <math>p</math> is a positive integer and <math>x</math> and <math>y</math> are real numbers,
  
 
<math>x^{p+1}-y^{p+1}=(x-y)(x^p+x^{p-1}y+\cdots +xy^{p-1}+y^p)</math>
 
<math>x^{p+1}-y^{p+1}=(x-y)(x^p+x^{p-1}y+\cdots +xy^{p-1}+y^p)</math>
  
−
For example,
+
For example:
  
 
<math>x^2-y^2=(x-y)(x+y)</math>
 
<math>x^2-y^2=(x-y)(x+y)</math>
Line 17: Line 17:
 
<math>x^4-y^4=(x-y)(x^3+x^2y+xy^2+y^3)</math>
 
<math>x^4-y^4=(x-y)(x^3+x^2y+xy^2+y^3)</math>
  
−
Note that the number of terms in the ''long'' factor is equal to the exponent in the expression being factored.
+
Note that the number of terms in the second factor is equal to the exponent in the expression being factored.
  
−
An amazing thing happens when x and y differ by 1, say, x = y+1. Then x-y = 1 and  
+
An amazing thing happens when <math>x</math> and <math>y</math> differ by <math>1</math>, say, <math>x = y+1</math>. Then <math>x-y = 1</math> and  
  
 
<math>x^{p+1}-y^{p+1}=(y+1)^{p+1}-y^{p+1}</math>
 
<math>x^{p+1}-y^{p+1}=(y+1)^{p+1}-y^{p+1}</math>
Line 25: Line 25:
 
<math>=(y+1)^p+(y+1)^{p-1}y+\cdots +(y+1)y^{p-1} +y^p</math>.
 
<math>=(y+1)^p+(y+1)^{p-1}y+\cdots +(y+1)y^{p-1} +y^p</math>.
  
−
For example,
+
For example:
  
 
<math>(y+1)^2-y^2=(y+1)+y</math>
 
<math>(y+1)^2-y^2=(y+1)+y</math>
Line 33: Line 33:
 
<math>(y+1)^4-y^4=(y+1)^3+(y+1)^2y+(y+1)y^2+y^3</math>
 
<math>(y+1)^4-y^4=(y+1)^3+(y+1)^2y+(y+1)y^2+y^3</math>
  
−
If we also know that <math>y\geq 0</math> then
+
If we also know that <math>y\geq 0</math> then:
  
 
<math>2y\leq (y+1)^2-y^2\leq 2(y+1)</math>
 
<math>2y\leq (y+1)^2-y^2\leq 2(y+1)</math>
Line 42: Line 42:
  
 
<math>(p+1)y^p\leq (y+1)^{p+1}-y^{p+1}\leq (p+1)(y+1)^p</math>
 
<math>(p+1)y^p\leq (y+1)^{p+1}-y^{p+1}\leq (p+1)(y+1)^p</math>
 +
 +
==Sum of Cubes==
 +
<math>1^3=1^2</math>
 +
 +
<math>1^3+2^3 =3^2</math>
 +
 +
<math>1^3 +2^3+3^3=6^2</math>
 +
 +
<math>1^3+2^3+3^3+4^3=10^2</math>
 +
 +
<math>1^3+2^3+3^3+4^3+\ldots+n^3=\left(\frac{n(n+1)}{2} \right) ^2=(1+2+3+4+\ldots+n)^2</math>
 +
 +
==Factorizations of Sums of Powers==
 +
<math>x^2-y^2=(x-y)(x+y)</math>
 +
 +
<math>x^3-y^3=(x-y)(x^2+xy+y^2)</math>
 +
 +
<math>x^4-y^4=(x-y)(x^3+x^2y+xy^2+y^3)</math>
 +
 +
Note that all these sums of powers can be factorized as follows:
 +
 +
If we have a difference of powers of degree <math>n</math>, then
 +
 +
<cmath>x^n-y^n=(x-y)(x^{n - 1}+x^{n-2}y+x^{n-3}y^2...+y^{n - 1})</cmath>
 +
 +
Note, of course, that some of the signs simply change when we have sum of powers instead of difference. Regardless, observe that the "longer" polynomial in the factorization is simply a binomial theorem expansion of the binomial <math>(x+y)^n</math>, except for the fact that the coefficient on each of the terms is <math>1</math>. This can be quite useful in problems that might have a sum of powers expression as well as an application of the binomial theorem.
 +
 +
- icecreamrolls8
  
 
==See Also==
 
==See Also==
 
* [[Factoring]]
 
* [[Factoring]]
 
* [[Difference of squares]], an extremely common specific case of this.
 
* [[Difference of squares]], an extremely common specific case of this.
−
{{stub}}
+
* [[Binomial Theorem]]
−
[[Category:Elementary algebra]]
+
 
 +
[[Category:Algebra]]

Latest revision as of 05:25, 6 May 2024

The sum and difference of powers are powerful factoring techniques that, respectively, factor a sum or a difference of certain powers.

Sums of Odd Powers

$a^{2n+1}+b^{2n+1}=(a+b)(a^{2n}-a^{2n-1}b+a^{2n-2}b^2-\ldots-ab^{2n-1}+b^{2n})$

Differences of Powers

If $p$ is a positive integer and $x$ and $y$ are real numbers,

$x^{p+1}-y^{p+1}=(x-y)(x^p+x^{p-1}y+\cdots +xy^{p-1}+y^p)$

For example:

$x^2-y^2=(x-y)(x+y)$

$x^3-y^3=(x-y)(x^2+xy+y^2)$

$x^4-y^4=(x-y)(x^3+x^2y+xy^2+y^3)$

Note that the number of terms in the second factor is equal to the exponent in the expression being factored.

An amazing thing happens when $x$ and $y$ differ by $1$, say, $x = y+1$. Then $x-y = 1$ and

$x^{p+1}-y^{p+1}=(y+1)^{p+1}-y^{p+1}$

$=(y+1)^p+(y+1)^{p-1}y+\cdots +(y+1)y^{p-1} +y^p$.

For example:

$(y+1)^2-y^2=(y+1)+y$

$(y+1)^3-y^3=(y+1)^2+(y+1)y+y^2$

$(y+1)^4-y^4=(y+1)^3+(y+1)^2y+(y+1)y^2+y^3$

If we also know that $y\geq 0$ then:

$2y\leq (y+1)^2-y^2\leq 2(y+1)$

$3y^2\leq (y+1)^3-y^3\leq 3(y+1)^2$

$4y^3\leq (y+1)^4-y^4\leq 4(y+1)^3$

$(p+1)y^p\leq (y+1)^{p+1}-y^{p+1}\leq (p+1)(y+1)^p$

Sum of Cubes

$1^3=1^2$

$1^3+2^3 =3^2$

$1^3 +2^3+3^3=6^2$

$1^3+2^3+3^3+4^3=10^2$

$1^3+2^3+3^3+4^3+\ldots+n^3=\left(\frac{n(n+1)}{2} \right) ^2=(1+2+3+4+\ldots+n)^2$

Factorizations of Sums of Powers

$x^2-y^2=(x-y)(x+y)$

$x^3-y^3=(x-y)(x^2+xy+y^2)$

$x^4-y^4=(x-y)(x^3+x^2y+xy^2+y^3)$

Note that all these sums of powers can be factorized as follows:

If we have a difference of powers of degree $n$, then

\[x^n-y^n=(x-y)(x^{n - 1}+x^{n-2}y+x^{n-3}y^2...+y^{n - 1})\]

Note, of course, that some of the signs simply change when we have sum of powers instead of difference. Regardless, observe that the "longer" polynomial in the factorization is simply a binomial theorem expansion of the binomial $(x+y)^n$, except for the fact that the coefficient on each of the terms is $1$. This can be quite useful in problems that might have a sum of powers expression as well as an application of the binomial theorem.

- icecreamrolls8

See Also

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=Sum_and_difference_of_powers&oldid=219049"