Simplify $(2 p + 3 q - 4 r)^{2} + (2 p - 3 q - 4 r)^{2}$

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Solution

Let us first take $(2 p + 3 q - 4 r)^{2}$
$(2 p + 3 q - 4 r)^{2} = {2 p + 3 q + (- 4 r)}^{2}$
Using, $(a + b + c)^{2} = a^{2} + b^{2} + c^{2} + 2 a b + 2 b c + 2 a c$
$= (2 p)^{2} + (3 q)^{2} + (- 4 r)^{2} + 2 (2 p) (3 q) + 2 (3 q) (- 4 r) + 2 (- 4 r) (2 p)$
$= 4 p^{2} + 9 q^{2} + 16 r^{2} + 12 p q - 24 q r - 16 r p$ ...(i)
Now, let us solve,
$(2 p - 3 q - 4 r)^{2} = {2 p + (- 3 q) + (- 4 r)}^{2}$
Using, $(a + b + c)^{2} = a^{2} + b^{2} + c^{2} + 2 a b + 2 b c + 2 a c$
$= (2 p)^{2} + (- 3 q)^{2} + (- 4 r)^{2} + 2 (2 p) (- 3 q) + 2 (- 3 q) (- 4 r) + 2 (- 4 r) (2 p)$
$= 4 p^{2} + 9 q^{2} + 16 r^{2} - 12 p q + 24 q r - 16 r p$ .....(ii)
Adding the corresponding sides of (i)& (ii) we get,
$(2 p + 3 q - 4 r)^{2} + (2 p - 3 q - 4 r)^{2}$
$= 4 p^{2} + 9 q^{2} + 16 r^{2} + 12 p q - 24 q r - 16 r p + 4 p^{2} + 9 q^{2} + 16 r^{2} - 12 p q + 24 q r - 16 r p$
$= 8 p^{2} + 18 q^{2} + 32 r^{2} - 32 r p .$
$= 2 (4 p^{2} + 9 q^{2} + 16 r^{2} - 16 r p) .$

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