Question:
If \(x + \frac{1}{x} = \sqrt{3}\), what is the value of \(x^{18} + x^{12} + x^6 + 1\)?
यदि \(x + \frac{1}{x} = \sqrt{3}\) है, तो \(x^{18} + x^{12} + x^6 + 1\) का मान क्या है?
Options:
- 0
- 1
- 2
- -1
- 0
- 1
- 2
- -1
Correct Answer: a)
Explanation:
Given \(x + \frac{1}{x} = \sqrt{3}\).
Cubing both sides:
\(x^3 + \frac{1}{x^3} + 3\left(x + \frac{1}{x}\right) = 3\sqrt{3}\)
\(x^3 + \frac{1}{x^3} + 3\sqrt{3} = 3\sqrt{3} \implies x^3 + \frac{1}{x^3} = 0\).
Multiplying by \(x^3\):
\(x^6 + 1 = 0 \implies x^6 = -1\).
Now substitute into the expression:
\(x^{18} + x^{12} + x^6 + 1 = (x^6)^3 + (x^6)^2 + (x^6) + 1\)
\(= (-1)^3 + (-1)^2 + (-1) + 1 = -1 + 1 – 1 + 1 = 0\).
Cubing both sides:
\(x^3 + \frac{1}{x^3} + 3\left(x + \frac{1}{x}\right) = 3\sqrt{3}\)
\(x^3 + \frac{1}{x^3} + 3\sqrt{3} = 3\sqrt{3} \implies x^3 + \frac{1}{x^3} = 0\).
Multiplying by \(x^3\):
\(x^6 + 1 = 0 \implies x^6 = -1\).
Now substitute into the expression:
\(x^{18} + x^{12} + x^6 + 1 = (x^6)^3 + (x^6)^2 + (x^6) + 1\)
\(= (-1)^3 + (-1)^2 + (-1) + 1 = -1 + 1 – 1 + 1 = 0\).
\(x + \frac{1}{x} = \sqrt{3}\) का दोनों पक्षों में घन करने पर:
\(x^3 + \frac{1}{x^3} = 0 \implies x^6 = -1\)
अब व्यंजक में मान रखने पर:
\((x^6)^3 + (x^6)^2 + (x^6) + 1 = (-1)^3 + (-1)^2 + (-1) + 1 = -1 + 1 – 1 + 1 = 0\)।
\(x^3 + \frac{1}{x^3} = 0 \implies x^6 = -1\)
अब व्यंजक में मान रखने पर:
\((x^6)^3 + (x^6)^2 + (x^6) + 1 = (-1)^3 + (-1)^2 + (-1) + 1 = -1 + 1 – 1 + 1 = 0\)।