Multiply by \(x-1\) on both sides and divide by \(x+1\) on both sides. Also \(x\ne1,-1\)
The expression becomes,
\[\dfrac{x^{4}-1}{x^{4}+1}=\dfrac{x^{3}-1}{x^{3}+1}\]
Now cross multiplying, we have,
\[(x^{4}-1)(x^{3}+1)=(x^{4}+1)(x^{3}-1)\]
\[ x^{7}-x^{3}+x^{4}-1=x^{7}+x^{3}-x^{4}-1\]
\[\implies 2*(x^{3})(x-1)=0\]
But since \(x\ne1\), we have \(x=0\). But substituting \(x=0\) in the main equation we see its not satisfied. Therefore the above expression is not true for any \(x\) belonging to \(R\)

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TopNewestMultiply by \(x-1\) on both sides and divide by \(x+1\) on both sides. Also \(x\ne1,-1\) The expression becomes, \[\dfrac{x^{4}-1}{x^{4}+1}=\dfrac{x^{3}-1}{x^{3}+1}\] Now cross multiplying, we have, \[(x^{4}-1)(x^{3}+1)=(x^{4}+1)(x^{3}-1)\] \[ x^{7}-x^{3}+x^{4}-1=x^{7}+x^{3}-x^{4}-1\] \[\implies 2*(x^{3})(x-1)=0\] But since \(x\ne1\), we have \(x=0\). But substituting \(x=0\) in the main equation we see its not satisfied. Therefore the above expression is not true for any \(x\) belonging to \(R\)

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Thats correct @Vignesh S

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I got a notification you tagged me. Lol

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\( \frac{ x^3 + x^2 + x + 1}{x^3 - x^2 + x - 1} = \frac{x^2 + x + 1}{x^2 - x + 1} \)

\( \frac{(x^2 + 1)\cdot(x+1)}{(x^2+1)\cdot(x-1)} = \frac{x^2 + x + 1}{x^2 - x + 1} \)

Since \( x \in \Re , x^2 + 1 \neq 0\) \(\forall x \), so:

\( \frac{x+1}{x-1} = \frac{x^2 + x + 1}{x^2 - x + 1} \)

\( x^3 + 1 = x^3 - 1 \)

\(1 = -1\)

Which is false, regardless of \(x\)

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I did it as:

On LHS in the numerator and denominator I took common and cancelled things off..

Further I applied componendo and dividendo to obtain \(x^{2}\) = \(x^{2}\) + 1 which is not possible hence proved....

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