Write the following in the expand form:

Solution:

(i) We have,

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

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

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

(ii) We have,

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

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

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

$4 a^{2}+9 b^{2}+c^{2}-12 a b+6 b c-4 c a$

$\therefore(2 a-3 b-c)^{2}=4 x^{2}+9 y^{2}+c^{2}-12 a b+6 b c-4 c a$

(iii) We have,

$(-3 x+y+z)^{2}=\left[(-3 x)^{2}+y^{2}+z^{2}+2(-3 x) y+2 y z+2(-3 x) z\right.$

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

$9 x^{2}+y^{2}+z^{2}-6 x y+2 y z-6 x z$

$(-3 x+y+z)^{2}=9 x^{2}+y^{2}+z^{2}-6 x y+2 x y-6 x y$

(iv) We have,

$(m+2 n-5 p)^{2}=m^{2}+(2 n)^{2}+(-5 p)^{2}+2 m \times 2 n+(2 \times 2 n \times-5 p)+2 m \times-5 p$

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

$(m+2 n-5 p)^{2}=m^{2}+4 n^{2}+25 p^{2}+4 m n-20 n p-10 p m$

(v) We have,

$(2+x-2 y)^{2}=2^{2}+x^{2}+(-2 y)^{2}+2(2)(x)+2(x)(-2 y)+2(2)(-2 y)$

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

$=4+x^{2}+4 y^{2}+4 x-4 x y-8 y$

$(2+x-2 y)^{2}=4+x^{2}+4 y^{2}+4 x-4 x y-8 y$

(vi) We have,

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

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

$\left(a^{2}+b^{2}+c^{2}\right)^{2}=a^{4}+b^{4}+c^{4}+2 a^{2} b^{2}+2 b^{2} c^{2}+2 c^{2} a^{2}$

(vii) We have,

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

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

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

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

(viii) We have,

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

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

$\therefore\left(\frac{\mathrm{x}}{\mathrm{y}}+\frac{\mathrm{y}}{\mathrm{z}}+\frac{\mathrm{z}}{\mathrm{x}}\right)^{2}=\left(\frac{\mathrm{x}^{2}}{\mathrm{y}^{2}}\right)+\left(\frac{\mathrm{y}^{2}}{\mathrm{z}^{2}}\right)+\left(\frac{\mathrm{z}^{2}}{\mathrm{x}^{2}}\right)+2 \frac{\mathrm{x}}{\mathrm{z}}+2 \frac{\mathrm{y}}{\mathrm{x}}+2 \frac{\mathrm{x}}{\mathrm{y}}$

(ix) We have,

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

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

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

(x) We have,

$(x+2 y+4 z)^{2}=x^{2}+(2 y)^{2}+(4 z)^{2}+2 x \times 2 y+2 \times 2 y \times 4 z+2 x \times 4 z$

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

$(x+2 y+4 z)^{2}=x^{2}+4 y^{2}+16 z^{2}+4 x y+16 y z+8 x z$

(xi) We have,

$(2 x-y+z)^{2}=(2 x)^{2}+(-y)^{2}+(z)^{2}+2(2 x)(-y)+2(-y)(z)+2(2 x)(z)$

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

$(2 x-y+z)^{2}=4 x^{2}+y^{2}+z^{2}-4 x y-2 y z+4 x z$

(xii) We have,

$(-2 x+3 y+2 z)^{2}=(-2 x)^{2}+(3 y)^{2}+(2 z)^{2}+2(-2 x)(3 y)+2(3 y)(2 z)+2(-2 x)(2 z)$

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

$(-4 x+6 y+4 z)^{2}=4 x^{2}+9 y^{2}+4 z^{2}-12 x y+12 y z-8 x z$