If $\dfrac{x}{a} = \dfrac{y}{b} = \dfrac{z}{c}$, prove that :

(i) $\dfrac{x^{2} + y^{2} + z^{2}}{a^{2} + b^{2} + c^{2}} = \Big(\dfrac{px + qy + rz}{pa + qb + rc}\Big)^{2}$

(ii) $\dfrac{x^{3}}{a^{3}} + \dfrac{y^{3}}{b^{3}} + \dfrac{z^{3}}{c^{3}} = \dfrac{3xyz}{abc}$

(iii) $\dfrac{x^{3}}{a^{2}} + \dfrac{y^{3}}{b^{2}} + \dfrac{z^{3}}{c^{2}} = \dfrac{(x + y + z)^{3}}{(a + b + c)^{2}}$

(iv) $\dfrac{ax - by}{(a + b)(x - y)} + \dfrac{by - cz}{(b + c)(y - z)} + \dfrac{cz - ax}{(c + a)(z - x)} = 3$

(i) Given,

⇒ $\dfrac{x}{a} = \dfrac{y}{b} = \dfrac{z}{c}$ = k (let)

⇒ x = ak, y = bk, z = ck.

Substituting values of x, y and z in L.H.S. of the equation $\dfrac{x^{2} + y^{2} + z^{2}}{a^{2} + b^{2} + c^{2}} = \Big(\dfrac{px + qy + rz}{pa + qb + rc}\Big)^{2}$, we get :

$\Rightarrow \dfrac{x^2 + y^2 + z^2}{a^2 + b^2 + c^2} \\[1em] \Rightarrow \dfrac{k^2 a^2 + k^2 b^2 + k^2 c^2}{a^2 + b^2 + c^2} \\[1em] \Rightarrow \dfrac{k^2 (a^2 + b^2 + c^2)}{a^2 + b^2 + c^2} \\[1em] \Rightarrow k^2.$

Substituting values of x, y and z in R.H.S. of the equation $\dfrac{x^{2} + y^{2} + z^{2}}{a^{2} + b^{2} + c^{2}} = \Big(\dfrac{px + qy + rz}{pa + qb + rc}\Big)^{2}$, we get :

$\Rightarrow \Big(\dfrac{px + qy + rz}{pa + qb + rc}\Big)^2 \\[1em] \Rightarrow \Big(\dfrac{p(ka) + q(kb) + r(kc)}{pa + qb + rc}\Big)^2 \\[1em] \Rightarrow \Big(\dfrac{k(pa + qb + rc)}{pa + qb + rc}\Big)^2 \\[1em] \Rightarrow k^2.$

Since, L.H.S. = R.H.S.

Hence, proved that $\dfrac{x^{2} + y^{2} + z^{2}}{a^{2} + b^{2} + c^{2}} = \Big(\dfrac{px + qy + rz}{pa + qb + rc}\Big)^{2}$.

(ii) Given,

⇒ $\dfrac{x}{a} = \dfrac{y}{b} = \dfrac{z}{c}$ = k (let)

⇒ x = ak, y = bk, z = ck.

Substituting values of x, y and z in L.H.S. of the equation $\dfrac{x^{3}}{a^{3}} + \dfrac{y^{3}}{b^{3}} + \dfrac{z^{3}}{c^{3}} = \dfrac{3xyz}{abc}$, we get:

$\Rightarrow \dfrac{x^3}{a^3} + \dfrac{y^3}{b^3} + \dfrac{z^3}{c^3} \\[1em] \Rightarrow \dfrac{k^3 a^3}{a^3} + \dfrac{k^3 b^3}{b^3} + \dfrac{k^3 c^3}{c^3} \\[1em] \Rightarrow k^3 + k^3 + k^3 \\[1em] \Rightarrow 3k^3.$

Substituting values of x, y and z in R.H.S. of the equation $\dfrac{x^{3}}{a^{3}} + \dfrac{y^{3}}{b^{3}} + \dfrac{z^{3}}{c^{3}} = \dfrac{3xyz}{abc}$, we get:

$\Rightarrow \dfrac{3xyz}{abc} \\[1em] \Rightarrow \dfrac{3(ka)(kb)(kc)}{abc} \\[1em] \Rightarrow \dfrac{3k^3 abc}{abc} \\[1em] \Rightarrow 3k^3.$

Since, L.H.S. = R.H.S.

Hence, proved that $\dfrac{x^{3}}{a^{3}} + \dfrac{y^{3}}{b^{3}} + \dfrac{z^{3}}{c^{3}} = \dfrac{3xyz}{abc}$.

(iii) Given,

⇒ $\dfrac{x}{a} = \dfrac{y}{b} = \dfrac{z}{c}$ = k (let)

⇒ x = ak, y = bk, z = ck.

Substituting values of x, y and z in L.H.S. of the equation $\dfrac{x^{3}}{a^{2}} + \dfrac{y^{3}}{b^{2}} + \dfrac{z^{3}}{c^{2}} = \dfrac{(x + y + z)^{3}}{(a + b + c)^{2}}$, we get:

$\Rightarrow \dfrac{x^3}{a^2} + \dfrac{y^3}{b^2} + \dfrac{z^3}{c^2} \\[1em] \Rightarrow \dfrac{k^3 a^3}{a^2} + \dfrac{k^3 b^3}{b^2} + \dfrac{k^3 c^3}{c^2} \\[1em] \Rightarrow k^3 a + k^3 b + k^3 c \\[1em] \Rightarrow k^3 (a + b + c).$

Substituting values of x, y and z in R.H.S. of the equation $\dfrac{x^{3}}{a^{2}} + \dfrac{y^{3}}{b^{2}} + \dfrac{z^{3}}{c^{2}} = \dfrac{(x + y + z)^{3}}{(a + b + c)^{2}}$, we get:

$\Rightarrow \dfrac{(x + y + z)^3}{(a + b + c)^2} \\[1em] \Rightarrow \dfrac{(k(a + b + c))^3}{(a + b + c)^2} \\[1em] \Rightarrow \dfrac{k^3 (a + b + c)^3}{(a + b + c)^2} \\[1em] \Rightarrow k^3 (a + b + c).$

Since, L.H.S. = R.H.S.

Hence, proved that $\dfrac{x^{3}}{a^{2}} + \dfrac{y^{3}}{b^{2}} + \dfrac{z^{3}}{c^{2}} = \dfrac{(x + y + z)^{3}}{(a + b + c)^{2}}$.

(iv) Given,

⇒ $\dfrac{x}{a} = \dfrac{y}{b} = \dfrac{z}{c}$ = k (let)

⇒ x = ak, y = bk, z = ck.

Substituting values of x, y and z in $\dfrac{ax - by}{(a + b)(x - y)}$, we get:

$\Rightarrow \dfrac{ax - by}{(a + b)(x - y)} \\[1em] \Rightarrow \dfrac{a(ka) - b(kb)}{(a + b)(ka - kb)} \\[1em] \Rightarrow \dfrac{k(a^2 - b^2)}{k(a + b)(a - b)} \\[1em] \Rightarrow \dfrac{k(a - b)(a + b)}{k(a + b)(a - b)} \\[1em] \Rightarrow 1….(1)$

Substituting values of x, y and z in $\dfrac{by - cz}{(b + c)(y - z)}$,we get:

$\Rightarrow \dfrac{by - cz}{(b + c)(y - z)} \\[1em] \Rightarrow \dfrac{b(kb) - c(kc)}{(b + c)(kb - kc)} \\[1em] \Rightarrow \dfrac{k(b^2 - c^2)}{k(b + c)(b - c)} \\[1em] \Rightarrow \dfrac{k(b - c)(b + c)}{k(b + c)(b - c)} \\[1em] \Rightarrow 1….(2)$

Substituting values of x, y and z in $\dfrac{cz - ax}{(c + a)(z - x)}$,we get:

$\Rightarrow \dfrac{cz - ax}{(c + a)(z - x)} \\[1em] \Rightarrow \dfrac{c(kc) - a(ka)}{(c + a)(kc - ka)} \\[1em] \Rightarrow \dfrac{k(c^2 - a^2)}{k(c + a)(c - a)} \\[1em] \Rightarrow \dfrac{k(c - a)(c + a)}{k(c + a)(c - a)} \\[1em] \Rightarrow 1….(3)$

Adding 1,2 and 3 we get,

$\Rightarrow \dfrac{ax - by}{(a + b)(x - y)} + \dfrac{by - cz}{(b + c)(y - z)} + \dfrac{cz - ax}{(c + a)(z - x)} \\[1em] \Rightarrow 1 + 1 + 1 \\[1em] \Rightarrow 3.$

Since, L.H.S = R.H.S

Hence, proved that $\dfrac{ax - by}{(a + b)(x - y)} + \dfrac{by - cz}{(b + c)(y - z)} + \dfrac{cz - ax}{(c + a)(z - x)} = 3$.