Newton sums in easier notation

\[\LARGE a^n+b^n+c^n=(a+b+c)(a^{n-1}+b^{n-1}+c^{n-1})-(ab+bc+ac)(a^{n-2}+b^{n-2}+c^{n-2})+abc(a^{n-3}+b^{n-3}+c^{n-3})\]

$\huge a^n+b^n=(a+b)(a^{n-1}+b^{n-1})-ab(a^{n-2}+b^{n-2})$

$\huge a^n+\frac{1}{a^n}=(a+\frac{1}{a})(a^{n-1}+\frac{1}{a^{n-1}})-(a^{n-2}+\frac{1}{a^{n-2}})$

For exercises: Try this problem

Try this very nice set by Aditya.

And this

#Algebra #SymmetricDifference #Evaluation #Roots #NewtonSums

Easy Math Editor

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Comments

Can we use newton sums for more then three unknowns also . . I mean can we do this:- $(a^4+b^4+c^4+d^4)=(a+b+c+d)(a^3+b^3+c^3+d^3)-(ab+bc+bd+ac+cd+ad)(a^2+b^2+c^2+c^2)+abcd(a+b+c+d)$

Aman Sharma - 6 years, 7 months ago

$(a^n+b^n+c^n+d^n)=(a+b+c+d)(a^{n-1}+b^{n-1}+c^{n-1}+d^{n-1})-(ab+ac+ad+bc+bd+cd) (a^{n-2}+b^{n-2}+c^{n-2}+d^{n-2})+(abc+acd+bcd+abd) (a^{n-3}+b^{n-3}+c^{n-3}+d^{n-3})-abcd (a^{n-4}+b^{n-4}+c^{n-4}+d^{n-4})$