Question

If x+1x=√3, what is the value of x^30+x^24+x^18+x^12+x^6+1?

Answer 1

So we have,

$x+\dfrac {1}{x}=\sqrt3$

and we're given to find the value of

$x^{30}+x^{24}+x^{18}+x^{12}+x^6+1$

First consider the given equation.

$x+\dfrac {1}{x}=\sqrt3$

Taking the cubes of both the sides,

$\left (x+\dfrac {1}{x}\right)^3=\left(\sqrt3\right)^3\\\\\\x^3+3x+\dfrac {3}{x}+\dfrac {1}{x^3}=3\sqrt3\\\\\\x^3+\dfrac {1}{x^3}+3\left (x+\dfrac {1}{x}\right)=3\sqrt 3\\\\\\x^3+\dfrac {1}{x^3}+3\sqrt 3=3\sqrt 3\\\\\\x^3+\dfrac {1}{x^3}=0\\\\\\\dfrac {x^6+1}{x^3}=0\\\\\\x^6+1=0\\\\\\x^6=-1$

Then,

$x^{30}+x^{24}+x^{18}+x^{12}+x^6+1\\\\=(x^6)^5+(x^6)^4+(x^6)^3+(x^6)^2+x^6+1\\\\=(-1)^5+(-1)^4+(-1)^3+(-1)^2+-1+1\\\\=-1+1-1+1-1+1\\\\=\mathbf {0}$

Thus the answer is simply 0.

#answerwithquality

#BAL

Answer 2

Answer:

0

Step-by-step explanation:

Given,

(x + 1/x) = √3

On cubing both sides, we get

⇒ (x + 1/x)³ = (√3)³

⇒ x³ + 1/x³ + 3 * (x + 1/x) = 3√3

⇒ x³ + 1/x³ + 3 * √3 = 3√3

⇒ x³ + 1/x³ = 0

⇒ x³ = -1/x³

⇒ x⁶ = -1

Now,

x³⁰ + x²⁴ + x¹⁸ + x¹² + x⁶ + 1

= (x⁶)⁵ + (x⁶)⁴ + (x⁶)³ + (x⁶)² + (x⁶)¹ + 1

= (-1)⁵ + (-1)⁴ + (-1)³ + (-1)² + (-1)¹ + 1

= -1 + 1 - 1 + 1 - 1 + 1

= 0

Hope it helps!