Question

If a + b + c = 9 and ab +bc + ca = 26, find the value of a³ + b³+ c³ − 3abc

Answer 1

In the given problem, we have to find value of  `a^3 + b^3 + c^3 - 3abc`

Given  `a+b+c = 9, ab + bc + ca = 26`

We shall use the identity

`(a+b+c)^2 = a^2 + b^2 + 2 (ab + bc + ca)`

`(a+b+c)^2 = a^2 + b^2 + c^2 + 2(26)`

              `(9)^2 = a^2 + b^2 + c^2 + 52`

        `81 - 52 = a^2 b^ + c^2`

                 `29 = a^2 +b^2 + c^2`

We know that 

`a^3 + b^3 + c^3 - 3abc = (a+b+c)(a^2 + b^2 +c^2 - ab - bc - ca)`

`a^3 + b^3 + c^3 - 3abc = (a+b+c)[(a^2 + b^2 +c^2) -( ab + bc +ca)]`

Here substituting  `a+b + c = 9,ab + bc + ca = 26,a^2 + b^2 + c^2 = 29 ` we get, 

`a^3 + b^3 + c^3 - 3abc = 9 [(29 - 26)]`

                                   ` = 9 xx 3`

                                  ` = 27`

Hence the value of  `a^3 + b^3 + c^3 - 3abc` is  27.