Let's denote \(p(x) = x^{n}+y^{n}\). Due to remainder theorem (little Bezout's theorem) remainder will be \(p(-y)= (-y) ^{n} + y^{n}=0\), because n is odd. Therefore the remainder is 0, so p(x) is divisible by x+y.
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Karan Pedja
·
2 years, 3 months ago

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Hi,@Ayush Choubey
If n is odd, then \[x^{n}+y^{n}=(x+y)(x^{n-1}-x^{n-2}y+x^{n-3}y^{2}-....+y^{n-1}) \]
which can be proved easily by expanding or using induction.
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Abhijeet Verma
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1 year, 7 months ago

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TopNewestLet's denote \(p(x) = x^{n}+y^{n}\). Due to remainder theorem (little Bezout's theorem) remainder will be \(p(-y)= (-y) ^{n} + y^{n}=0\), because n is odd. Therefore the remainder is 0, so p(x) is divisible by x+y. – Karan Pedja · 2 years, 3 months ago

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Hi,@Ayush Choubey If n is odd, then \[x^{n}+y^{n}=(x+y)(x^{n-1}-x^{n-2}y+x^{n-3}y^{2}-....+y^{n-1}) \] which can be proved easily by expanding or using induction. – Abhijeet Verma · 1 year, 7 months ago

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