I-need-a-general-rule-for-factorising-1-a-n-b-n-2-a-n-b-n-when-n-is-even-and-when-n-is-odd-

Question Number 116163 by harckinwunmy last updated on 01/Oct/20

I need a general rule for factorising

1) a^{n} + b^{n}

2) a^{n} - b^{n}

when n is even and when n is odd .

Answered by 1549442205PVT last updated on 02/Oct/20

I need a general rule for factorising

1) a^{n} + b^{n}

2) a^{n} - b^{n}

when n is even and when n is odd .

i) when n is even : n = 2 k we have

1) a^{2 k} + b^{2 k} = {(a^{k} + b^{k})}^{2} - {2 a}^{k} b^{k}

(a^{k} + \sqrt{{2 a}^{k} b^{k}} + b^{k}) (a^{k} - \sqrt{{2 a}^{k} b^{k}} + b^{k})

2) a^{2 k} - b^{2 k} = (a^{k} + b^{k}) (a^{k} - b^{k})

= (a^{k} + b^{k}) (a - b) (a^{k - 1} + a^{k - 2} b + a^{k - 3} b^{2}

+ \dots + b^{k - 1})

ii) When n is odd : n = 2 k + 1 we have

1) a^{2 k + 1} + b^{2 k + 1} = (a + b) (a^{2 k} - a^{2 k - 1} b

+ a^{2 k - 2} b^{2} - \dots + b^{2 k})

2) a^{2 k + 1} - b^{2 k + 1} = (a - b) (a^{2 k} + a^{2 k - 1} b +

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