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Difference between revisions of "Sum and difference of powers"

(odd, it does not work for even)
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<math>1^3+2^3+3^3+4^3=10^2</math>
 
<math>1^3+2^3+3^3+4^3=10^2</math>
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==Factorizations of Sums of Powers==
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<math>x^2-y^2=(x-y)(x+y)</math>
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<math>x^3-y^3=(x-y)(x^2+xy+y^2)</math>
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<math>x^4-y^4=(x-y)(x^3+x^2y+xy^2+y^3)</math>
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Note that all these sums of powers can be factorized as follows:
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If we have a sum of powers of degree "n", then
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(x^n)±(y^n)=(x±y)(x^n+(x^(n-1))y+(x^(n-2))y^2...+y^n)
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 +
Note 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.
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- icecreamrolls8
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==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.
 +
* [[Binomial Theorem]]
 
{{stub}}
 
{{stub}}
 
[[Category:Elementary algebra]]
 
[[Category:Elementary algebra]]

Revision as of 14:12, 5 May 2021

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 long 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

$1^3=1^2$

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

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

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

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 sum of powers of degree "n", then

(x^n)±(y^n)=(x±y)(x^n+(x^(n-1))y+(x^(n-2))y^2...+y^n)

Note 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

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