Question : Simplify the expression: $\frac{(u–v)^{3}+(v–w)^{3}+(w–u)^{3}}{(u^{2}–v^{2})^{3}+(v^{2}–w^{2})^{3}+(w^{2}–u^{2})^{3}}$
Option 1: $\frac{1}{(u+v)(v+w)(w+u)}$
Option 2: $1$
Option 3: $\frac{3}{(u+v)(v+w)(w+u)}$
Option 4: $0$
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Correct Answer: $\frac{1}{(u+v)(v+w)(w+u)}$
Solution :
Given: $\frac{(u–v)^{3}+(v–w)^{3}+(w–u)^{3}}{(u^{2}–v^{2})^{3}+(v^{2}–w^{2})^{3}+(w^{2}–u^{2})^{3}}$
We know that, when $x+y+z=0$, then $x^3+y^3+z^3=3xyz$
So, $(a–b)^{3}+(b–c)^{3}+(c–a)^{3}=3(a–b)(b–c)(c–a)$
Applying this to the given expression, we have,
$\frac{3(u–v)(v–w)(w–u)}{3(u^{2}–v^{2})(v^{2}–w^{2})(w^{2}–u^{2})}$
= $\frac{3(u–v)(v–w)(w–u)}{3(u–v)(u+v)(v–w)(v+w)(w–u)(w+u)}$
= $\frac{1}{(u+v)(v+w)(w+u)}$
Hence, the correct answer is $\frac{1}{(u+v)(v+w)(w+u)}$.
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