Question

If \(x^2+x+1=0\), then the value of \(\left(x + \frac{1}{x}\right)^2 + \left(x^2 + \frac{1}{x^2}\right)^2 + \left(x^3 + \frac{1}{x^3}\right)^2 + \dots + \left(x^{25} + \frac{1}{x^{25}}\right)^2\) is:

• A. 128
• B. 175
• C. 145
• D. 162

Hint:

Whenever you see the equation \(x^2+x+1=0\), immediately think of the complex cube roots of unity, \(\omega\) and \(\omega^2\).
Their properties (\(\omega^3=1, 1+\omega+\omega^2=0\)) are fundamental to solving such problems quickly.
The value of \(x^n + 1/x^n\) follows a simple periodic pattern.

Answer 1

The Correct Option is C

To solve this problem, we are given that \(x^2 + x + 1 = 0\). We need to evaluate the expression \((x + \frac{1}{x})^2 + (x^2 + \frac{1}{x^2})^2 + \cdots + (x^{25} + \frac{1}{x^{25}})^2\).

First, let's find the roots of the equation \(x^2 + x + 1 = 0\). Using the quadratic formula, \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), with \(a = 1\), \(b = 1\), \(c = 1\), we have:

\(x = \frac{-1 \pm \sqrt{1 - 4}}{2} = \frac{-1 \pm \sqrt{-3}}{2} = \frac{-1 \pm \sqrt{3}i}{2}\) 

The roots of the equation are complex conjugates: \(\omega = \frac{-1 + \sqrt{3}i}{2}\) and \(\omega^2 = \frac{-1 - \sqrt{3}i}{2}\). It is known that for these roots, they satisfy \(\omega^3 = 1\). Therefore, \(\omega^n\) will cycle every three terms: \(\omega^1 = \omega\), \(\omega^2\), \(\omega^3 = 1\), \(\omega^4 = \omega\), and so on.

Now, we continue by finding the value of \((x + \frac{1}{x})\). With \(x = \omega\) or \(\omega^2\), and knowing \(\omega^3 = 1\), we have:

\(x + \frac{1}{x} = \omega + \bar{\omega} = -1\) (where \(\bar{\omega}\) is the complex conjugate)

Thus, \((x + \frac{1}{x})^2 = (-1)^2 = 1\). Using the identity \((x + \frac{1}{x})^2 = x^2 + \frac{1}{x^2} + 2\), we find:

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

Now, we calculate \((x^2 + \frac{1}{x^2})^2 = (-1)^2 = 1\).

Using similar computation for \((x^3 + \frac{1}{x^3})\) we get:

\(x^3 + \frac{1}{x^3} = (x + \frac{1}{x})(x^2 + \frac{1}{x^2}) - (x + \frac{1}{x}) = (-1)(-1) - (-1) = 1 + 1 = 2\)

Thus, \((x^3 + \frac{1}{x^3})^2 = 2^2 = 4\).

Similarly, if any term has a power \(\equiv 0 \ (\text{mod } 3)\), their sum is \(x^3 + \frac{1}{x^3} = 2\).

Notice the repetitive pattern due to \(\omega^3 = 1\), and calculate the complete sum of squares:

• For each \(n \equiv 0 \ (\text{mod } 3)\), \((x^n + \frac{1}{x^n})^2 = 4\)
• For each \(n \not\equiv 0 \ (\text{mod } 3)\), \((x^n + \frac{1}{x^n})^2 = 1\)

Counting the terms in powers of 3 from 3 to 24 (3, 6, 9, ..., 24), we have 8 terms where it is 4, and 17 terms at 1.

The summation is \(8 \times 4 + 17 \times 1 = 32 + 17 = 49\).

Thus, the total value of the expression is \(145\).

Therefore, the correct answer is \(145\).