Big Numbers

if ${ x }^{ 2 }+x+1=0$

what is the value of

$5{ x }^{ 234 }-{ x }^{ 99 }$

#Algebra

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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}`	$\boxed{123}$

Comments

Solving the first equation, we get

$x = \frac{-1 \pm \sqrt{3}i}{2}$

Let us denote $\omega = \frac{-1 + \sqrt{3}i}{2}$

Then it is easy to see that

$\omega^2 = \left(\frac{-1 + \sqrt{3}i}{2}\right)^2 = \frac{-1 - \sqrt{3}i}{2}$

Thus, the two roots of the equation are $\omega,\ \omega^2$

These are well-known by the name of cube roots of unity because when you solve the equation $x^3 - 1=0$ , you get the roots as $1,\ \omega,\ \omega^2$ . And thus, by Vieta's formula, we have

$1+\omega+\omega^2=0$

$\text{\&} \qquad \omega^3 = 1$

Using the above properties, we can easily find that -

$5x^{234}-x^{99} = 5\omega^{234} - \omega^{99} = 5(\omega^3)^{78} - (\omega^3)^{33} = 5-1 = \boxed{4}$ -

Also, it doesn't matters whichever value of $x$ you substitute, you'll always get the same answer as you can see -

$5x^{234}-x^{99} = 5(\omega^2)^{234} - (\omega^2)^{99} = 5(\omega^3)^{156} - (\omega^3)^{66} = 5-1 = \boxed{4}$

I hope you got my method.

Thanks.

Kishlaya Jaiswal - 6 years, 2 months ago

good method kishlaya.thanks.but i know this problem has a very simpler way.unfortunately i couldnt remember it .could anyone solve this problem in another way???

mobin moradi - 6 years, 2 months ago

Roots of the given equation are $\Large{w~\&~w^2}$ . Hence the answer is 4.

Aniket Verma - 6 years, 2 months ago

could you please explain more about it

safa m - 6 years, 2 months 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}`	$\boxed{123}$