If x : a = y : b, prove that,

$\dfrac{x^4 + a^4}{x^3 + a^3} + \dfrac{y^4 + b^4}{y^3 + b^3} = \dfrac{(x + y)^4 + (a + b)^4}{(x + y)^3 + (a + b)^3}.$

Given, x : a = y : b,

Let, $\dfrac{x}{a}$ = $\dfrac{y}{b}$ = k

∴ x = ak, y = bk

Putting values of x and y in L.H.S. first,

$= \dfrac{(ak)^4 + a^4}{(ak)^3 + a^3} + \dfrac{(bk)^4 + b^4}{(bk)^3 + b^3} \\[1em] = \dfrac{a^4(k^4 + 1)}{a^3(k^3 + 1)} + \dfrac{b^4(k^4 + 1)}{b^3(k^3 + 1)} \\[1em] = \dfrac{a(k^4 + 1) + b(k^4 + 1)}{k^3 + 1} \\[1em] = \dfrac{(a + b)(k^4 + 1)}{k^3 + 1}$

Now, putting values in R.H.S., $= \dfrac{(x + y)^4 + (a + b)^4}{(x + y)^3 + (a + b)^3} \\[1em] = \dfrac{(ak + bk)^4 + (a + b)^4}{(ak + bk)^3 + (a + b)^3} \\[1em] = \dfrac{k^4(a + b)^4 + (a + b)^4}{k^3(a + b)^3 + (a + b)^3} \\[1em] = \dfrac{(a + b)^4(k^4 + 1)}{(a + b)^3(k^3 + 1)} \\[1em] = \dfrac{(a + b)(k^4 + 1)}{k^3 + 1}.$

Since, L.H.S. = R.H.S. Hence, proved that,

$\dfrac{x^4 + a^4}{x^3 + a^3} + \dfrac{y^4 + b^4}{y^3 + b^3} = \dfrac{(x + y)^4 + (a + b)^4}{(x + y)^3 + (a + b)^3}.$