a+b=10, a^2+b^2=167, a^3+b^3=? Помогите решить.

2ab=(a+b)^2-(a^2+b^2)=100-167=-67; ab=-67/2;
a^3+b^3=(a+b)(a^2+b^2-ab)=10*(167+67/2)=2005
Ответ: 2005.

а + b = 10 (a = 10 - b)
(a+b)^2 = 100
a^2 + 2ab + b^2 = 100 ( a^2 + b^2 = 167)
167 + 2ab = 100
2ab = -67
2(10-b)*b = -67
(20-2b)*b = -67
20b - 2b^2 + 67 = 0
-2b^2 + 20b + 67 = 0
D = 400 + 536 = 936
b1 = -20 - √(936) / -4 = (20+√936)/4 = 5 + 1.5√26
b2 = -20 + √(936) / -4 = 5 - 1.5√26

a + b = 10
a + 5 - 1.5√26 = 10
a2 = 10 - 5 + 1.5√26 = 5 + 1.5√26
a1 = 10 - 5 - 1.5√26 = 5 - 1.5√26

(a1 + b1) = 5 - 1.5√26 + 5 + 1.5√26 = 10
(a2 + b2) = 5 + 1.5√26 + 5 - 1.5√26 = 10

a^2 + b^2 = (5-1.5√26)^2 + (5+1.5√26)^2 = 25 - 3√26 + 58.5 + 25 + 3√26 + 58.5 = 50 + 117 = 167 ( совпадает, а2 и b2 сами проверите)

a^3 + b^3 = 125 - 15√26 + 292.5 - 37.5√26 + 117 - 87.75√26 + 125 + 15√26 + 292.5 + 37.5√26 + 117 + 87.87√26 = 125 + 125 + 292.5 + 292.5 + 117 + 117 = 1069

бигус