# Question #02d02

Jun 28, 2017

$2 {a}^{2} + 2 {b}^{2} + 2 {c}^{2} + 4 b c$

#### Explanation:

${\left[a + b + c\right]}^{2} + {\left[a - b - c\right]}^{2}$

Solution

Expand the brackets..

$\left[a + b + c\right] \left[a + b + c\right] + \left[a - b - c\right] \left[a - b - c\right]$

$\left[{a}^{2} + a b + a c + a b + {b}^{2} + b c + a c + b c + {c}^{2}\right] + \left[{a}^{2} - a b - a c - a b + {b}^{2} + b c - a c + b c + {c}^{2}\right]$

Collecting like terms...

$\left[{a}^{2} + {b}^{2} + {c}^{2} + \left(a b + a b\right) + \left(a c + a c\right) + \left(b c + b c\right)\right] + \left[{a}^{2} + {b}^{2} + {c}^{2} + \left(- a b - a b\right) + \left(- a c - a c\right) + \left(b c + b c\right)\right]$

$\left[{a}^{2} + {b}^{2} + {c}^{2} + 2 a b + 2 a c + 2 b c\right] + \left[{a}^{2} + {b}^{2} + {c}^{2} - 2 a b - 2 a c + 2 b c\right]$

Collect like terms in both brackets...

$\left({a}^{2} + {a}^{2}\right) + \left({b}^{2} + {b}^{2}\right) + \left({c}^{2} + {c}^{2}\right) + \cancel{2 a b - 2 a b} + \cancel{2 a c - 2 a c} + \left(2 b c + 2 b c\right)$

$\to$ $2 {a}^{2} + 2 {b}^{2} + 2 {c}^{2} + 4 b c$