Question

find value of [119]²+[119][111]+[111]²/[119]³-[111]³

Answer 1

Let 119 = a and 111 = b

$then, \frac{119^2+119.111+111^2}{119^3-111^3}=\frac{a^2+ab+b^2}{a^3-b^3}$
we know,
a³ - b³ = (a - b)(a² + ab + b²) use this here,

$\frac{a^2+ab+b^2}{a^3-b^3}=\frac{a^2+ab+b^2}{(a-b)(a^2+ab+b^2)} \\ \\ = \frac{1}{(a - b)} \\ \\ = \frac{1}{119 - 111} \\ \\ = \frac{1}{8}$

Answer 2

Heya folk !!

Here's the answer you are looking for.

$\frac{ {(119)}^{2} + (119)(111) + {(111)}^{2} }{ {(119)}^{3} - {(111)}^{3} }$


Multiply, (119 - 111) to both numerator and denominator, so we get,

$\frac{ {[(119)}^{2} + (119)(111) + {(111)}^{2}] (119 - 111)}{ [{(119)}^{3} - {(111)}^{3}] (119 - 111) }$


The numerator is in the form of
(a - b)(a² + ab + b²) = a³ - b³

So, it becomes,

$\frac{ [{119}^{3} - {111}^{3}] }{[{119}^{3} - {111}^{3}](119 - 111)} \\ \\ = \frac{1}{(119 - 111)} \\ \\ = \frac{1}{8} (ans)$

★★ HOPE THAT HELPS ☺️ ★★