Difference between revisions of "Sum and difference of powers"

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<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+2^3 =3^2</math>
<math>1^3 +2^3+3^3=6^2</math>
==See Also==
==See Also==

Revision as of 11:08, 6 April 2016

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

Sums of 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


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

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

See Also

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