6=a+b+c
11=a*b+(a+b)*c
6=abc
And other ones with more variables such as
10=a+b+c+d
35=a*b+(a+b)*c+(a+b+c)*d
50=(abc)+d((a*b)(a+b)*c)
24=abcd
So that I can get the roots of higher degree polynomials without having to approximate.
Thanks.
6=a+b+c
11=a*b+(a+b)*c
6=abc
And other ones with more variables such as
10=a+b+c+d
35=a*b+(a+b)*c+(a+b+c)*d
50=(abc)+d((a*b)(a+b)*c)
24=abcd
So that I can get the roots of higher degree polynomials without having to approximate.
Thanks.
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Roots of $$ x^3 - 6 x^2 + 11x-6, $$ $$ x^4 - 10x^3 + 35 x^2 - 50 x + 24. $$ Which seem to be $(1,2,3)$ and $(1,2,3,4)$
SEE VIETE