Question

If a+b+c=6 and a2+b2+c2=14 and a3+b3+c3=36 find value of abc

Answer 1

Answer-

(a+b+c)^2=a2+b2+c2+2(ab+bc+ca)
(6)2=14+2(ab+bc+ca)
36-14=2(ab+bc+ca)
22/2=ab+bc+ca
ab+bc+ca=11

a^3+b^3+c^3-3abc
=(a+b+c)(a2+b2+c2-ab-bc-ca)
=(6)(14-11)
=(6)(3)
=18

Since, a^3+b^3+c^3-3abc=18
36-3abc=18
-3abc=18-36
-3abc=-18
abc=6

Hope it helps

Answer 2

Given : a+b+c=6 ,   a² + b² + c²  = 14 , a³ + b³ + c³ = 36

To find : Value of abc

Solution:

a+b+c=6

Squaring both sides

=> a² + b² + c²  + 2(ab + bc + ac) = 36

=> 14 + 2(ab + bc + ac) = 36

=> 2(ab + bc + ac) = 22

=> ab + bc + ac = 11

a³ + b³ + c³ – 3abc = (a + b + c) (a² + b² + c² – ab – bc – ca)

=> 36 - 3abc = 6 ( 14 - 11)

=> 36 - 3abc = 18

=> 3abc = 18

=> abc = 6

Additional info

a , b & c = 1  , 2 & 3   ( can be one of the solution)

Learn more:

a³ + b³ + c³ -3abc = ( a + b + c)(a² + b² + c² - ab - bc - ca) .

https://brainly.in/question/1195178