Use algebraic identities to expand the following algebraic equations.

Solution:

(i) We have,

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

$\left[\because(x+y+z)^{2}=x^{2}+y^{2}+z^{2}+2 x y+2 y z+2 x z\right]$

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

$(a+b+c)^{2}+(a-b+c)^{2}=2 a^{2}+2 b^{2}+2 c^{2}+4 c a$

(ii)  We have,

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

$\left[\because(x+y+z)^{2}=x^{2}+y^{2}+z^{2}+2 x y+2 y z+2 x z\right]$

$\left.=a^{2}+b^{2}+c^{2}+2 a b+2 b c+2 c a-a^{2}-b^{2}-c^{2}+2 a b+2 b c-2 c a\right)$

$=4 a b+4 b c$

$(a+b+c)^{2}-(a-b+c)^{2}=4 a b+4 b c$

(iii) We have,

$(a+b+c)^{2}+(a+b-c)^{2}+(a+b-c)^{2}$

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

$\left[\because(x+y+z)^{2}=x^{2}+y^{2}+z^{2}+2 x y+2 y z+2 x z\right]$

$=3 a^{2}+3 b^{2}+3 c^{2}+2 a b+2 b c+2 c a-2 b c-2 a b-2 c a-2 b c+2 a b$

$=3 x^{2}+3 y^{2}+3 z^{2}+2 a b-2 b c+2 c a$

$(a+b+c)^{2}+(a+b-c)^{2}+(a-b+c)^{2}=3 a^{2}+3 b^{2}+3 c^{2}+2 a b-2 b c+2 c a$

$(a+b+c)^{2}+(a+b-c)^{2}+(a-b+c)^{2}=3\left(a^{2}+b^{2}+c^{2}\right)+2(a b-b c+c a)$

(iv) We have,

$(2 x+p-c)^{2}-(2 x-p+c)^{2}$

$=\left[2 x^{2}+p^{2}+(-c)^{2}+2(2 x) p+2 p(-c)+2(2 x)(-c)\right]-\left[4 x^{2}+(-p)^{2}+c^{2}+2(2 x)(-p)+2 c(-p)+2(2 x) c\right]$

$\left[\because(x+y+z)^{2}=x^{2}+y^{2}+z^{2}+2 x y+2 y z+2 x z\right]$

$(2 x+p-c)^{2}-(2 x-p+c)^{2}=\left[4 x^{2}+p^{2}+c^{2}+4 x p-2 p c-4 x c\right]-\left[4 x^{2}+p^{2}+c^{2}-4 x p-2 p c+4 x c\right]$

Opening the bracket,

$\left.(2 x+p-c)^{2}-(2 x-p+c)^{2}=4 x^{2}+p^{2}+c^{2}+4 x p-2 p c-4 c x-4 x^{2}-p^{2}-c^{2}+4 x p+2 p c-4 c x\right]$

$(2 x+p-c)^{2}-(2 x-p+c)^{2}=8 x p-8 x c$

$=8 x(p-c)$

Hence, $(2 x+p-c)^{2}-(2 x-p+c)^{2}=8 x(p-c)$

(v) We have,

$\left(x^{2}+y^{2}+(-z)^{2}\right)^{2}-\left(x^{2}(-y)^{2}+z^{2}\right)^{2}$

$=\left[x^{4}+y^{4}+(-z)^{4}+2 x^{2} y^{2}+2 y^{2}(-z)^{2}+2 x^{2}(-z)^{2}\right]-\left[x^{4}+(-y)^{4}+z^{4}-2 x^{2} y^{2}-2 y^{2} z^{2}+2 x^{2} z^{2}\right]$

$\left[\because(x+y+z)^{2}=x^{2}+y^{2}+z^{2}+2 x y+2 y z+2 x z\right]$

Taking the negative sign inside,

$=\left[x^{4}+y^{4}+(-z)^{4}+2 x^{2} y^{2}+2 y^{2}(-z)^{2}+2 x^{2}(-z)^{2}\right]-\left[x^{4}+(-y)^{4}+z^{4}-2 x^{2} y^{2}-2 y^{2} z^{2}+2 x^{2} z^{2}\right]$

$=4 x^{2} y^{2}-4 z^{2} x^{2}$

Hence, $\left(x^{2}+y^{2}+(-z)^{2}\right)^{2}-\left(x^{2}(-y)^{2}+z^{2}\right)^{2}=4 x^{2} y^{2}-4 z^{2} x^{2}$