A Nice Polynomial System | #algebra #polynomials

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a+b+c=10 and a^3+b^3+c^3=3abc+10. ab+ac+bc=?
#systemsofequations #algebra #algebraicexpressions
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[XJWill]

Or just use the Newton-Girard identities for symmetric power polynomials, specifically these two:
{1} p3 = p2*e1 - p1*e2 + 3*e3 = p2*e1 - e1*e2 + 3*e3
{2} p2 = p1*e1 - 2*e2 = e1^2 - 2*e2
Combined with the given two equations:
e1 = 10
p3 = 10 + 3*e3
Just substitute these last two, as well as {2}, into {1} to eliminate e1, p2, p3:
10 + 3*e3 = (10^2 - 2*e2)*10 - 10*e2 + 3*e3
10 = 1000 - 20*e2 - 10*e2
e2 = 33

[andy_in_colorado]

I chose to start with a + b + c = 10 and cube it, ultimately through expanding and substituting arriving at 30(ab + ac + bc) = 990, hence ab + ac + bc = 33:

a^3 + b^3 + c^3 = 10 + 3abc
a + b + c = 10
ab + ac + bc = ?
(a + b + c)^3 = [a + (b + c)]^3 = a^3 + (b + c)^3 + 3a(b + c)(a + b + c)
= a^3 + b^3 + c^3 + 3bc(b + c) + (3ab + 3ac)(a + b + c)
= a^3 + b^3 + c^3 + 3(b^2)c + 3bc^2 + 3(a^2)b + 3ab^2 + 3abc + 3(a^2)c + 3abc + 3ac^2
= a^3 + b^3 + c^3 + 3(a^2)b + 3ab^2 + 3(a^2)c + 3ac^2 + 3(b^2)c + 3bc^2 + 6abc
= 10 + 3abc + 3ab(a + b) + 3ac(a + c) + 3bc(b + c) + 6abc
= 10 + 9abc + 3ab(10 - c) + 3ac(10 - b) + 3bc(10 - a)
= 10 + 9abc + 30ab - 3abc + 30ac - 3abc + 30bc - 3abc
= 10 + 30(ab + ac + bc) = 10^3 = 1000
30(ab + ac + bc) = 990
ab + ac + bc = 33

In other words, "no pain, no gain". At least it worked!

[sohampine]

Bro when are you going to upload in the aplusbi channel?, btw great video😊

[samarthchohan]

a^3+b^3+c^3-3abc is a very famous identity actually so when I saw that I immediately factored

[mbmillermo]

Very clever! (Especially that initial factorization.)

[luygvkz]

Can you do some mutli variable calculus problems bring in some concepts like green thermos

[carlyet]

Interesting your algebra is anti-computational.
Processing the general instead of reducing the specific.
Exploring by generation:
a+b+c=10 & (a+b+c)^2 and (a+b+c)^3

[yoav]

Nice. Do'nt be afraid of talking too much, words do'nt cost much😃

[loflight]

xyz = ? If ab + bc + ca = 33, xyz = 0 in my calculation...