Let's Evaluate An Algebraic Expression

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[ashokchaudhary]

Both methods are really good.
Still, When x^2 = Root(3)*x -1
x^6 = (x^2)^3 = 3*Root(3)*x^3 -3Root(3)*x*(Root(3)*x -1) -1
Or x^6 = 3*Root(3)*x^3 -3Root(3)*x*( x^2 ) -1 = 3*Root(3)*x^3 -3Root(3)*x^3 -1
Or x^6 = -1
Rest is (x^6)^3 + (x^6)^2 + 1 = -1+1+1 --> 1

[danieldepaula]

I think the quickest way maybe is:

x + (1/x) = sqrt(3)
[x + (1/x)]³ = [sqrt(3)]³
x³ + (1/x³) + 3x + (3/x) = 3sqrt(3)
x³ + (1/x³) + 3*[x + (1/x)] = 3sqrt(3)
x³ + (1/x³) = 0

(x^18) + (x^12) + 1 = ?
(x^18) + (x^12) = ? - 1
(x^15)*[x³ + (1/x³)] = ? - 1
(x^15)*0 = ? - 1
? = 1

[SidneiMV]

x + 1/x = √3
x² + 1/x² = 1
x⁴ = x² - 1
x⁸ = x⁴ + 1 - 2x² = (x² - 1) + 1 - 2x²
x⁸ = -x²
x⁶ = - 1 => x¹² = 1

x¹⁸ + x¹² + 1 = x¹²(x⁶ + 1) + 1
x¹⁸ + x¹² + 1 = x¹²(-1 + 1) + 1

*x¹⁸ + x¹² + 1 = 1*

[visweswararoach]

Just cubing on both sides and simplify gives x^6 = - 1 and hence the required result equal to 1.

[htdgtigidog]

hephep hooray
x¹⁸ + x¹² + 1 = (x¹⁵)(x³ + 1/x³) + 1

Squaring x + 1/x = sqrt(3) gives x² + 1/x² = 1. Then,
x³ + 1/x³ = (x + 1/x)(x² + 1/x² - 1) = sqrt(3)(1 - 1) = 0.

So, x¹⁸ + x¹² + 1 = (x¹⁵)(x³ + 1/x³) + 1 = 1

[Lemda_gtr]

Don’t be teacher😂😂, u skipped a critical steps