Question

if a+b+c=1, ab +bc +ca= _1 and abc =_1 then find a cube + b cube + c cube ​

Answer 1

Answer:

given a+b+c=1,ab+bc+ca=-1,abc=-1

a³+b³+c³=?

we have formlua (a+b+c)³=a³+b³+c³+3a²b+3a²c+3b²a+3b²c+3c²a+3c²b+6abc

(a+b+c)³=a³+b³+c³+3[a²(b+c)+b²(a+c)+c²(a+b)]+6abc

(a+b+c)³=a³+b³+c³+3[a²(1-a)+b²(1-b)+c²(1-c)]+6abc

:.(a+b+c=1)

(a+b+c)³=a³+b³+c³+3(a²-a³+b²-b³+c²-c³)+6abc

(a+b+c)³=a³+b³+c³-3(a³+b³+c³)+3(a²+b²+c²)+6abc

(a+b+c)³=-2(a³+b³+c³)+3[(a+b+c)²-2(ab+bc+ca)]+6abc

:.(a+b+c)²=a²+b²+c²+2ab+2ac+2bc

2(a³+b³+c³)=3[(a+b+c)²-2(ab+bc+ca)]-(a+b+c)³

substituting given values (in question) now

2(a³+b³+c³)=3[1²-2(-1)]-1³

2(a³+b³+c³)=3(1+2)-1

2(a³+b³+c³)=3(3)-1

2(a³+b³+c³)=9-1

2(a³+b³+c³)=8

a³+b³+c³=8/2

a³+b³+c³=4

answer is 4