Question

If a+b+c=6, a^2+b^2+c^2=14, a^3+b^3+c^3=36, find abc

Answer 1

A+b+c=6........................................1

a²+b²+c²=14.................................2

a³+b³+c³=36.................................3

We know that,
(a+b+c)²=a²+b²+c²+2ab+2bc+2ac

(a+b+c)²=a²+b²+c²+2(ab+bc+ac)

By putting eq1 and eq2 we get,

(6)²=14+2(ab+bc+ac)


(36-14)/2=ab+bc+ac

ab+bc+ac=11................................4


By multiplying eq1 and eq4 we get,

(a+b+c)(ab+bc+ac)=6×11

a²b+abc+a²c+b²a+b²c+abc+abc+c²b+c²a=66

a²b+a²c+b²a+b²c+c²a+c²b+3abc=66

a²b+a²c+b²a+b²c+c²a+c²b+2abc+abc=66

a²b+a²c+b²c+b²a+c²a+c²b+2ab=66-abc..................................................5


We also know that,

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


Putting eq1 and eq3 we get,


(6)³=36+3{a²b+a²c+b²a+b²c+c²a+c²b+2abc}

(216-36)=3{a²b+a²c+b²a+b²c+c²a+c²b+2abc}

60=a²b+a²c+b²a+b²c+c²a+c²b+2ab

putting eq5 in above equation we get,

60=66-abc

abc=66-60

abc=6