If $a\ =\ 2$, $b\ =\ 3$  and  $c\ =\ 1$, find the value of:
$a^2\ +\ b^2\ +\ -\ 2ab\ -\ 2bc\ -\ 2ca +\ 3abc$

Given: $a\ =\ 2$, $b\ =\ 3$  and  $c\ =\ 1$.


To find: Here we have to find the value of  $a^2\ +\ b^2\ +\ c^2 -\ 2ab\ -\ 2bc\ -\ 2ca +\ 3abc$.


Solution:

$a^2\ +\ b^2\ +\ c^2 -\ 2ab\ -\ 2bc\ -\ 2ca\ +\ 3abc$

Putting values of a, b and c:

$=\ 2^2\ +\ 3^2\ +\ 1^2 -\ 2(2)(3)\ -\ 2(3)(1)\ -\ 2(1)(2)\ +\ 3(2)(3)(1)$

$=\ 4\ +\ 9\ +\ 1 -\ 12\ -\ 6\ -\ 4\ +\ 18$

$=\ 14 -\ 22\ +\ 18$

$=\ 14 -\ 4$

$=$  10


So, the value is 10.

Updated on: 10-Oct-2022

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