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| − | If p is a positive integer and x and y are real numbers,
| + | #REDIRECT[[Sum and difference of powers]] |
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| − | <math>x^{p+1}-y^{p+1}=(x-y)(x^p+x^{p-1}y+\cdots +xy^{p-1}+y^p)</math>
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| − | For example,
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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 the number of terms in the ''long'' factor is equal to the exponent in the expression being factored.
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| − | An amazing thing happens when x and y differ by 1, say, x = y+1. Then x-y = 1 and
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| − | <math>x^{p+1}-y^{p+1}=(y+1)^{p+1}-y^{p+1}</math>
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| − | <math>=(y+1)^p+(y+1)^{p-1}y+\cdots +(y+1)y^{p-1} +y^p</math>.
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| − | For example,
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| − | <math>(y+1)^2-y^2=(y+1)+y</math>
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| − | <math>(y+1)^3-y^3=(y+1)^2+(y+1)y+y^2</math>
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| − | <math>(y+1)^4-y^4=(y+1)^3+(y+1)^2y+(y+1)y^2+y^3</math>
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| − | If we also know that <math>y\geq 0</math> then
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| − | <math>2y\leq (y+1)^2-y^2\leq 2(y+1)</math>
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| − | <math>3y^2\leq (y+1)^3-y^3\leq 3(y+1)^2</math>
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| − | <math>4y^3\leq (y+1)^4-y^4\leq 4(y+1)^3</math>
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| − | <math>(p+1)y^p\leq (y+1)^{p+1}-y^{p+1}\leq (p+1)(y+1)^p</math>
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