I don' think questions should be given names.

Algebra Level 3

If x + 1 x = 3 x+\frac { 1 }{ x } =\sqrt { 3 } , find the value of x 18 + x 12 + x 6 + 1 { x }^{ 18 }+{ x }^{ 12 }+{ x }^{ 6 }+1 .


The answer is 0.

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Vaibhav Prasad by Vaibhav Prasad , 21, India

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7 solutions

Manish Dash
May 8, 2015

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Moderator note:

Yes, well done.

Nice method you got there. Here's another way to look at it:

Take the required expression as S S . From the given equation, we can see that x ≠ 0 , 1 x\neq 0,1 . So, we can regroup our required expression and use finite GP sum formula.

S = x 18 + x 12 + x 6 + 1 = ∑ r = 0 3 ( x 6 ) r = ( x 6 ) 4 − 1 x 6 − 1 S=x^{18}+x^{12}+x^6+1=\sum_{r=0}^3(x^6)^r=\frac{(x^6)^4-1}{x^6-1}

Back to our given equation, cube both sides and with a bit of manipulation, you'll get that,

x 3 + 1 x 3 = 0 ⟹ x 6 + 1 = 0 ⟹ x 6 = ( − 1 ) x^3+\frac{1}{x^3}=0\implies x^6+1=0\implies x^6=(-1)

Use this value of x 6 x^6 to evaluate S S as follows:

S = ( − 1 ) 4 − 1 ( − 1 ) − 1 = 1 − 1 − 2 = 0 S=\frac{(-1)^4-1}{(-1)-1}=\frac{1-1}{-2}=\boxed{0}

Prasun Biswas - 6 years, 1 month ago

I believe there is an error in this method. Specifically, you rely on this line for your solution:

x 6 + 1 x 6 = ( x 2 + 1 x 2 ) ( x 2 + 1 x 2 − 1 ) x^6 + \frac{1}{x^6} = (x^2 + \frac{1}{x^2})(x^2 + \frac{1}{x^2} - 1)

A quick glance shows that the leading term for the result of the right is actually x 4 x^4 , not x 6 x^6 . Working it through completely, you will find that:

( x 2 + 1 x 2 ) ( x 2 + 1 x 2 − 1 ) = x 4 + 1 x 4 − ( x 2 + 1 x 2 ) + 2 = x 4 + 1 x 4 + 1 (x^2 + \frac{1}{x^2})(x^2 + \frac{1}{x^2} - 1) = x^4 + \frac{1}{x^4} - (x^2 + \frac{1}{x^2}) + 2 = x^4 + \frac{1}{x^4}+1

And

x 6 + 1 x 6 ≠ x 4 + 1 x 4 + 1 x^6 + \frac{1}{x^6} \ne x^4 + \frac{1}{x^4}+1

But you do have a good start. Try this out:

x + 1 x = x 2 + 1 x = 3 = > ( x 2 + 1 ) 2 x 2 = 3 = x 4 + 2 x 2 + 1 x 2 = x 2 + 2 + 1 x 2 = 3 x 2 + 1 x 2 = 1 x + \frac{1}{x} = \frac{x^2+1}{x} = \sqrt{3} => \frac{(x^2+1)^2}{x^2} = 3 \\ = \frac{x^4+2x^2+1}{x^2} = x^2 + 2 + \frac{1}{x^2} = 3 \\ x^2 + \frac{1}{x^2} = 1

Next we need x 4 + 1 x 4 x^4 + \frac{1}{x^4} :

1 2 = ( x 2 + 1 x 2 ) 2 = ( x 4 + 1 x 2 ) 2 = x 4 + 2 + 1 x 4 = 1 = > x 4 + 1 x 4 = − 1 1^2 = (x^2 + \frac{1}{x^2})^2 = (\frac{x^4+ 1}{x^2})^2 \\ = x^4 + 2 + \frac{1}{x^4} = 1 \\ => x^4 + \frac{1}{x^4} = -1

Now:

x 2 ( x 4 + 1 x 4 ) = ( − 1 ) ( x 2 ) = > x 6 + 1 x 2 = ( − 1 ) ( x 2 ) = > x 6 = ( − 1 ) ( x 2 + 1 x 2 ) = > x 6 = ( − 1 ) ( 1 ) = − 1 x^2(x^4 + \frac{1}{x^4}) = (-1)(x^2) \\ => x^6 + \frac{1}{x^2} = (-1)(x^2) \\ => x^6 = (-1)(x^2 + \frac{1}{x^2}) \\ => x^6 = (-1)(1) = -1

Now to plug into our equation: x 18 + x 12 + x 6 + 1 = ( x 6 ) 3 + ( x 6 ) 2 + x 6 + 1 = ( − 1 ) 3 + ( − 1 ) 2 + ( − 1 ) + 1 = − 1 + 1 − 1 + 1 = 0 ∴ x 18 + x 12 + x 6 + 1 = 0 x^{18} + x^{12} + x^6 + 1 = {(x^6)}^3 + {(x^6)}^2 + x^6 + 1 \\ = (-1)^3 + (-1)^2 + (-1) + 1 = -1 + 1 - 1 + 1 = 0 \\ \therefore x^{18} + x^{12} + x^6 + 1 = 0

Quintessence Anx - 6 years, 1 month ago
Anandhu Raj
May 7, 2015

Given, x + 1 x = 3 x+\frac { 1 }{ x } =\sqrt { 3 }

Solving for x x , we get that x = 3 2 ± i 2 x=\frac { \sqrt { 3 } }{ 2 } \pm \frac { i }{ 2 }

Multiplying through out by i i ,

⇒ i x = i 3 2 ± − 1 2 \Rightarrow ix=\frac { i\sqrt { 3 } }{ 2 } \pm \frac { -1 }{ 2 }

⇒ i x = ω o r − ω \Rightarrow ix=\omega \quad or\quad -\omega

⇒ x = ω i o r − ω i \Rightarrow x=\frac { \omega }{ i } \quad or\quad -\frac { \omega }{ i }

Substituting the values of x x in x 18 + x 12 + x 6 + 1 { x }^{ 18 }+{ x }^{ 12 }+{ x }^{ 6 }+1 , we get,

− 1 + 1 − 1 + 1 = 0 -1+1-1+1=\boxed{0}

Note: ω i s t h e c u b e r o o t o f u n i t y \omega \quad is\quad the\quad cube\quad root\quad of\quad unity and i = − 1 i=\sqrt { -1 } .

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Moderator note:

Good work!

@challenge master, Didn't get your point....Would you please explain? How could x x be cube root of unity?

Anandhu Raj - 6 years, 1 month ago

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Sorry, typo. I misinterpreted the question. Well done!

Brilliant Mathematics Staff - 6 years, 1 month ago

Uhmm, @Calvin Lin Sir, There is a typo in the Challenge Master Note

Mehul Arora - 6 years, 1 month ago
Rajdeep Dhingra
May 8, 2015

x + 1 x = 3 x 2 − 3 x + 1 = 0 x = 3 2 ± i 1 2 We can take any value ⇒ x = cos ⁡ ( 30 ) + i sin ⁡ ( 30 ) x + \frac{1}{x} = \sqrt{3} \\ x^2 - \sqrt{3}x + 1 = 0 \\ x = \dfrac{\sqrt{3}}{2} \pm i\dfrac{1}{2} \\ \text{We can take any value} \\ \Rightarrow x = \cos(30) + i\sin(30)

Now We will apply De Moivre's Theorem to evaluate

x 18 + x 12 + x 6 + 1 = ( cos ⁡ ( 30 ) + i sin ⁡ ( 30 ) ) 18 + ( cos ⁡ ( 30 ) + i sin ⁡ ( 30 ) ) 12 + ( cos ⁡ ( 30 ) + i sin ⁡ ( 30 ) ) 6 + 1 Applying theorem we get = cos ⁡ ( 18 × 30 ) + i sin ⁡ ( 18 × 30 ) + cos ⁡ ( 12 × 30 ) + i sin ⁡ ( 12 × 30 ) + cos ⁡ ( 6 × 30 ) + i sin ⁡ ( 6 × 30 ) + 1 = − 1 + 0 + 1 + 0 − 1 + 0 + 1 0 x^{18} + x^{12} + x^6 + 1 \\ = (\cos(30) + i\sin(30))^{18} + (\cos(30) + i\sin(30))^{12} +\\ ( \cos(30) + i\sin(30))^6 + 1 \\ \text{Applying theorem we get} \\ = \cos(18\times30) + i\sin(18\times30) + \cos(12\times30) + i\sin(12\times30) + \\ \cos(6\times30) + i\sin(6\times30) + 1 \\ = -1 + 0 + 1 + 0 -1 + 0 + 1 \\ \boxed{0}

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Moderator note:

You can have a slight improvement. Hint: all the powers of x x are a multiple of 6 6 .

Javier Sanjuan
May 8, 2015

x 18 + x 12 + x 6 + 1 = ( x 12 + 1 ) ( x 6 + 1 ) = x 9 ( x 6 + 1 x 6 ) ( x 3 + 1 x 3 ) Then, given: x + 1 x = 3 x 2 + 1 x = 3 ( x 2 + 1 ) 3 x 3 = 3 3 x 6 + 3 x 4 + 3 x 2 + 1 x 3 = 3 3 x 3 + 3 x + 3 x + 1 x 3 = 3 3 x 3 + 1 x 3 = 0 ⇒ x 18 + x 12 + x 6 + 1 = 0 x^{18}+x^{12}+x^6+1=(x^{12}+1)(x^6+1)=x^9(x^6+\frac{1}{x^6})(x^3+\frac{1}{x^3}) \\ \text{Then, given:}\\x+\frac{1}{x}=\sqrt{3} \\ \frac{x^2+1}{x}=\sqrt{3}\\ \frac{(x^2+1)^3}{x^3}=3\sqrt{3} \\ \frac{x^6+3x^4+3x^2+1}{x^3}=3\sqrt{3}\\ x^3+3x+\frac{3}{x}+\frac{1}{x^3}=3\sqrt{3} \\ x^3+\frac{1}{x^3}=0 \\ \Rightarrow x^{18}+x^{12}+x^6+1=0

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Moderator note:

It's easier to start with cubing the equation x + 1 x = 3 x + \frac 1x = \sqrt 3 . Most people opted for the complex number approach, and you didn't. Nicely done!

x + 1 x = 3 . ∴ x ≠ 0. ( x + 1 x ) 3 = x 3 + 1 x 3 + 3 ∗ 1 ∗ ( x + 1 x ) . ⟹ x 3 + 1 x 3 + 3 ∗ 1 ∗ 3 = 3 3 . ∴ x 3 + 1 x 3 = 0. ∴ x 6 = − 1 , x 12 = 1 , x 18 = − 1. x+\dfrac 1 x =\sqrt 3. ~~\therefore x\neq0.\\\left (x+\dfrac 1 x \right )^3=x^3+\dfrac 1 {x^3}+3*1*( x+\dfrac 1 x).\\\implies~x^3+\dfrac 1 {x^3} +3*1*\sqrt 3=3\sqrt 3 .\\ \therefore~{x^3}+\dfrac 1 {x^3}=0.\\\therefore~x^6=-1,~~~~x^{12}=1,~~~x^{18}=-1.\\ ∴ x 18 + x 12 + x 6 + 1 = − 1 + 1 − 1 + 1 = 0 \therefore x^{18}+x^{12}+x^6+1=-1+1-1+1= ~~~~~~~~~~~~\color{#D61F06}{\Large 0}

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Hem Shailabh Sahu
May 10, 2015

We are given : x + 1 x = 3 . . . ( 1 ) x+\frac{1}{x}=\sqrt{3}...(1) Required expression : x 18 + x 12 + x 6 + 1 = ( x 6 + 1 ) ( x 12 + 1 ) = x 3 ( x 3 + 1 x 3 ) ( x 12 + 1 ) . . . ( 2 ) x^{18}+x^{12}+x^6+1=(x^6+1)(x^{12}+1)=x^{3}(x^3+\frac{1}{x^3})(x^{12}+1)...(2) Cubing both the sides of (1), ( x + 1 x ) 3 = ( 3 ) 3 (x+\frac{1}{x})^{3}=(\sqrt{3})^{3} ⇒ x 3 + 1 x 3 + 3 ( x ) ( 1 x ) ( x + 1 x ) = 3 3 \Rightarrow x^3+\frac{1}{x^3}+3(x)(\frac{1}{x})(x+\frac{1}{x})=3\sqrt{3} Using (1), ⇒ x 3 + 1 x 3 = 0... ( 3 ) \Rightarrow x^3+\frac{1}{x^3}=0...(3) Therefore, using (3), the value of required expression (2) = x 3 ( x 3 + 1 x 3 ) ( x 12 + 1 ) = 0 =x^{3}(x^3+\frac{1}{x^3})(x^{12}+1)=\boxed{0}

:D

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L N
May 10, 2015

x + 1 x = 3 x + \frac{1}{x} = \sqrt{3}

x 2 + 1 = 3 x x^2 + 1 = \sqrt{3}x

( x 2 + 1 ) 2 = 3 x 2 (x^2 + 1)^2 = 3x^2

x 4 + 2 x 2 + 1 = 3 x 2 x^4 + 2x^2 + 1 = 3x^2

x 4 − x 2 + 1 = 0 x^4 - x^2 + 1 = 0

Let z = x 2 z = x^2

z 2 − z + 1 = 0 z^2 - z + 1 = 0

z = ( − 1 ) 1 3 z = (-1)^{\frac{1}{3}}

x = ( − 1 ) 1 6 x = (-1)^{\frac{1}{6}}

There are other solutions for z z and x x , however this one does give a real valued solution as needed. Plug in the value of x x just found to get 0 0

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