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# Big Numbers

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

what is the value of

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

??

2 years, 8 months ago

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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.

- 2 years, 8 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???

- 2 years, 8 months ago

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

- 2 years, 8 months ago