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 $x\in \mathbb{R}$ .

#Algebra

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Comments

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 $R$

Vignesh S - 5 years, 1 month ago

Thats correct @Vignesh S

Nihar Mahajan - 5 years, 1 month ago

I got a notification you tagged me. Lol

A Former Brilliant Member - 5 years, 1 month ago

@A Former Brilliant Member – I first accidentally tagged you and then realized it was other vignesh :P

Nihar Mahajan - 5 years, 1 month ago

@Nihar Mahajan – Haha

A Former Brilliant Member - 5 years, 1 month 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 $x$

Guilherme Niedu - 5 years, 1 month 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....

Ankit Kumar Jain - 5 years, 1 month ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$