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# Fundamentals

$\dfrac{x^3+x^2+x+1}{x^3-x^2+x-1}=\dfrac{x^2+x+1}{x^2-x+1}$

Show that above is not true for any $$x\in \mathbb{R}$$.

Note by Akshat Sharda
11 months ago

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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 $$R$$ · 11 months ago

Thats correct @Vignesh S · 11 months ago

I got a notification you tagged me. Lol · 11 months ago

I first accidentally tagged you and then realized it was other vignesh :P · 11 months ago

Haha · 11 months ago

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

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

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 $$x$$ · 11 months ago

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.... · 11 months ago