Question

a+b+c=14
a^2+b^2+c^2=74
a^3+b^3+c^3=434
Find a,b,c

Answer 1

Given a + b + c = 14.

a^2 + b^2 + c^2 = 74

a^3 + b^3 + c^3 = 434.


We know that (a + b + c)^2 = a^2 + b^2 + c^2 + 2(ab + bc + ca)

                         (14)^2 = 74 + 2(ab + bc + ca)

                         196 - 74 = 2(ab + bc + ca)

                        122/2 = (ab + bc + ca).

                        61 = (ab + bc + ca)



We know that a^3 + b^3 + c^3 - 3abc = (a + b + c)(a^2 + b^2 + c^2 - ab - bc - ca)

                             434 - 3abc = (14)(74 - 61)

                             434 -3abc = 182

                             - 3abc = -252

                                abc = 84.


Hope this helps!

Answer 2

a+b+c=14
a+b=14-c

squaring both sides
a^2+b^2+2ab=196+c^2-28c
74-c^2+2ab=196+c^2-28c
2c^2-28c+270=2ab
c^2-14c+135=ab

first taking the third equation now,,
a^3+b^3+c^3=434
(a+b)(a^2+b^2-ab)=434-c^3
(14-c)(74-c^2-c^2+14c-135)=434-c^3
(14-c)(-1)(2c^2-14c+61)=434-c^3


now u need just to solve the equation and fund the answer