Would the graders want this to be in the same form as the one above, or would this be fine?

I think this problem has post in Batic Way.

hrithikguy wrote: Would the graders want this to be in the same form as the one above, or would this be fine? In other words, "Hey dear grader, it's your turn to simplify and give me full marks!"

Why should a grader care whether denominators have been cleared? It's a correct expression, and it satisfies the necessary condition that it contains only $k$ and no other instances of $x$ or $y$.

Valentin Vornicu wrote: Let $\frac{x^2+y^2}{x^2-y^2} + \frac{x^2-y^2}{x^2+y^2} = k$. Compute the following expression in terms of $k$: \[ E(x,y) = \frac{x^8 + y^8}{x^8-y^8} - \frac{ x^8-y^8}{x^8+y^8}. \] Ciprus Using the identities : $\boxed{\boxed{(A+B)^2+(A-B)^2=2(A^2+B^2),\ (A+B)^2-(A-B)^2=4AB}}$, $k=\frac{(x^2+y^2)^2+(x^2-y^2)^2}{(x^2+y^2)(x^2-y^2)}=\frac{2(x^4+y^4)}{x^4-y^4}\ \cdots [*]$ $\therefore E(x,\ y)=\frac{(x^8+y^8)^2-(x^8-y^8)^2}{(x^8+y^8)(x^8-y^8)}=\frac{4x^8y^8}{x^{16}-y^{16}}$ If $x\neq 0$, then set $t=\left(\frac{y}{x}\right)^4$, from $[*]$, we have $k=\frac{2\left\{1+\left(\frac{y}{x}\right)^4\right\}}{1-\left(\frac{y}{x}\right)^4}=\frac{2(1+t)}{1-t}\Longleftrightarrow t=\frac{k-2}{k+2}\ \because x\neq 0\Longrightarrow k\neq -2.$ $\therefore E(x,\ y)=\frac{4t^2}{1-t^4}=\frac{4\left(\frac{k-2}{k+2}\right)^2}{1-\left(\frac{k-2}{k+2}\right)^4}=\frac{4(k+2)^2(k-2)^2}{(k+2)^4-(k-2)^4}$ $=\frac{4(k+2)^2(k-2)^2}{\{(k+2)^2-(k-2)^2\}\{(k+2)^2+(k-2)^2\}}=\frac{4(k+2)^2(k-2)^2}{4\cdot k\cdot 2*2(k^2+2^2)}$ $=\boxed{\frac{(k+2)^2(k-2)^2}{4k(k^2+4)}}$ If $x=0$, then $k=-2$, yielding $E(0,\ y)=0.$ If $y=0$, then $k=2$, yielding $E(x,\ 0)=0.$

My solution : Lemma I used : if $\frac{a}{b} = \frac{c}{d} \text{then} \frac{a+b}{a-b} = \frac{c+d}{c-d} $ this lemma is also know'n as componendo and dividendo and the proof of this lemma is very simple. So $ \frac{x^2+y^2}{x^2-y^2}+\frac{x^2-y^2}{x^2+y^2}= k $ Simplifying gives $\frac{k}{2}=\frac{x^4+y^4}{x^4-y^4}$ Using Lemma $\frac{2x^4}{2y^4} = \frac{k+2}{k-2} $ $(\frac{x}{y})^4 = \frac{k+2}{k-2} $ Now we have to find $\frac{x^8+y^8}{x^8-y^8}+\frac{x^8-y^8}{x^8+y^8} $ which is equivalent to $ 2\frac{x^{16}+y^{16}}{x^{16}-y^{16}} $ $ 2\times\frac{(\frac{k+2}{k-2})^4+1}{(\frac{k+2}{k-2})^4-1}$ $2\times\frac{(k+2)^4+(k-2)^4}{(k+2)^4-(k-2)^4} $

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PP. Let $\frac{x^2+y^2}{x^2-y^2} + \frac{x^2-y^2}{x^2+y^2} = k$ . Compute the expression $E(x,y) = \frac{x^8 + y^8}{x^8-y^8} - \frac{ x^8-y^8}{x^8+y^8}$ in terms of $k$ . Proof. Since the expressions $\frac{x^2+y^2}{x^2-y^2} + \frac{x^2-y^2}{x^2+y^2}$ and $E(x,y)$ are homogeneously w.r.t. $\{x,y\}$ can use the substitution $\frac xy=t\ :$ $k=\frac {t^2+1}{t^2-1}+\frac {t^2-1}{t^2+1}\implies$ $k=\frac {2\left(t^4+1\right)}{t^4-1}\implies$ $\boxed{t^4=\frac {k+2}{k-2} }$ . Thus, $E= \frac{t^8 + 1}{t^8-1} - \frac{ t^8-1}{t^8+1}\implies$ $E=\frac {4t^8}{\left(t^8-1\right)\left(t^8+1\right)}$ . Therefore, $E=\frac {4\left(\frac {k+2}{k-2}\right)^2}{\left[\left(\frac {k+2}{k-2}\right)^2-1\right]\left[\left(\frac {k+2}{k-2}\right)^2+1\right]}=$ $\frac {4(k+2)^2(k-2)^2}{\left[(k+2)^2-(k-2)^2\right]\left[(k+2)^2+(k-2)^2\right]}\implies$ $\boxed{E=\frac {\left(k^2-4\right)^2}{4k\left(k^2+4\right)}}$ .

Let us denote $x^2:=a$ and $y^2:=b$. We want to find $\tfrac{a^4+b^4}{a^4-b^4}-\tfrac{a^4-b^4}{a^4+b^4}$. Note that \begin{align*}\frac{k}{2}+\frac{2}{k}=\frac{(a^2+b^2)^2+(a^2-b^2)^2}{a^4-b^4} & \iff \frac{k^2+4}{4k}=\frac{a^4+b^4}{a^4-b^4}.\end{align*} Then, \[\frac{k^2+4}{4k}-\frac{4k}{k^2+4}=\frac{a^4+b^4}{a^4-b^4}-\frac{a^4-b^4}{a^4+b^4}.\] Therefore, \[E(x,y)=\frac{k^2+4}{4k}-\frac{4k}{k^2+4}=\boxed{\frac{(k^2-4)^2}{4k(k^2+4)}}.\]

JBL wrote: Why should a grader care whether denominators have been cleared? It's a correct expression, and it satisfies the necessary condition that it contains only $k$ and no other instances of $x$ or $y$. This is not a fair statement IMO, considering how many graders do not accept fractions that are not in lowest terms or surds that are not simplified (for example)

djmathman, April 2012 wrote: We can finally say that $\dfrac{k^2+4}{4k}+\dfrac{4k}{k^2+4}=\dfrac{x^8+y^8}{x^8-y^8}+\dfrac{x^8-y^8}{x^8+y^8}$, which is our desired. Oops. Well now I know why this solution was wrong....

just calculation with no brain needed $k^4-8k^2+16/4k^3+16k$