Question

If a + b + c = 9 and ab + bc + ca =23, then a³ + b³ + c³ − 3abc =

पर्याय

• 108

• 207

• 669

• 729

Answer 1

We have to find the value of   `a^3 +b^3 +c^3 - 3abc`

Given  `a+b+c = 9,ab +bc +ca = 23`

Using identity   `(a+b+c)^2 = a^2 +b^2 +c^2 +2ab +2bc + 2ca` we get,

`(9)^2 = a^2 +b^2 +c^2 +2 (ab+bc +ca)`

` 9 xx 9 = a^2 +b^2 +c^2 +2 xx 23`

`81 = a^2 +b^2 +c^2 +46`

By transposing +46 to left hand side we get,

`81-46 = a^2 +b^2 +c^2`

          `35 = a^2 +b^2 +c^2`

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

 `9 xx [35 -23]`

` = 9 xx 12`

` = 108`

The value of  `a^3 +b^3 +c^3 -3abc` is 108.