Question

Find the square root of (a+b+c)^2+(a+b-c)^2+2(c^2-a^2-b^2-2ab)

Answer 1

$\sqrt{ {(a + b + c)}^{2} + {(a + b - c)}^{2} + 2( {c}^{2} - {a}^{2} - {b}^{2} - 2ab }$     We will expand both using identity (a+b+c)^2= a^2+b^2+c^2+2ab+2bc+2ac     $\sqrt{ {a}^{2} + {b}^{2} + {c}^{2} + 2ab + 2bc + 2ac + {a}^{2} + {b}^{2} + {c}^{2} + 2ab - 2bc - 2ac + 2( {c}^{2} - {a}^{2} - {b}^{2} - 2ab)}$     Now opening the brackets we will get     $\sqrt{ {a}^{2} + {b}^{2} + {c}^{2} + 2ab + 2bc + 2ac + {a}^{2} + {b}^{2} + {c}^{2} \: + 2ab - 2bc - 2ac + 2 {c}^{2} - 2 {a}^{2} - 2 {b}^{2} - 4ab}$     Now simplyfying this , we get :     $\sqrt{4 {c}^{2}}$     $\bold{Answer :}$     $\{2c}$

Answer 2

Hello!

We have:

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

Solution:

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

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

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

Note¹: let's simplify: $\left(a+b+c\right)\left(a+b+c\right)$

$aa+ab+ac+ab+bb+bc+ac+bc+cc$

$a^2+2ab+2ac+b^2+2bc+c^2$

Note²: let's simplify: $\left(a+b-c\right)\left(a+b-c\right)$

$aa+ab+a\left(-c\right)+ba+bb+b\left(-c\right)+\left(-c\right)a+\left(-c\right)b+\left(-c\right)\left(-c\right)$

$aa+ab-ac+ab+bb-bc-ac-bc+cc$

$aa+2ab-2ac+bb-2bc+cc$

$a^2+2ab-2ac+b^2-2bc+c^2$

Note³: let's simplify: $2\left(c^2-a^2-b^2-2ab\right)$

$2c^2+2\left(-a^2\right)+2\left(-b^2\right)+2\left(-2ab\right)$

$2c^2-2a^2-2b^2-4ab$

Now we have the replacement:

$\sqrt{a^2+2ab+2ac+b^2+2bc+c^2+a^2+2ab-2ac+b^2-2bc+c^2+2c^2-2a^2-2b^2-4ab}$

group the similar ones, let's see:

$\sqrt{a^2+a^2-2a^2+2ab+2ac+2ab-2ac-4ab+b^2+b^2-2b^2+2bc-2bc+c^2+c^2+2c^2}$

$\sqrt{\diagup\!\!\!\!\!\!2a^2-\diagup\!\!\!\!\!\!2a^2+\diagup\!\!\!\!\!\!4ab-\diagup\!\!\!\!\!\!4ab+\diagup\!\!\!\!\!\!2ac-\diagup\!\!\!\!\!\!2ac+\diagup\!\!\!\!\!\!2b^2-\diagup\!\!\!\!\!\!2b^2+\diagup\!\!\!\!\!\!2bc-\diagup\!\!\!\!\!\!2bc+2c^2+2c^2}$

$\sqrt{2c^2+2c^2}$

$\sqrt{4c^2}$

$\sqrt{4}\sqrt{c^2}$

$\boxed{\boxed{Answer = 2c}}\end{array}}\qquad\checkmark$

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I Hope this helps, greetings ... Dexteright02 ! =)