Sum and difference of powers

Revision as of 14:58, 5 May 2021 by Icecreamrolls8 (talk | contribs) (Factorizations of Sums of Powers)

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


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:




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


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

For example:




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


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

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


Factorizations of Sums of Powers




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

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


where a, b, c,... are equivalent to n-1, n-2, n-3,..., respectively.

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

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