The problem asks us to evaluate a summation involving the roots of the quadratic equation \(x^2 + x + 1 = 0\). Let the roots be \(\alpha\) and \(\beta\). The summation required is \(\mathop \sum \limits_{j = 0}^3 \left( {{\alpha ^j} + {\beta ^j}} \right)\).

Understanding the Roots of \(x^2 + x + 1 = 0\)

The given equation is a standard quadratic equation. We can find its roots using the quadratic formula \(x = \frac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}\). For \(x^2 + x + 1 = 0\), we have \(a=1\), \(b=1\), and \(c=1\).

The discriminant is \(\Delta = b^2 - 4ac = 1^2 - 4(1)(1) = 1 - 4 = -3\). Since the discriminant is negative, the roots are complex numbers.

The roots are:

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

These roots are the complex cube roots of unity, commonly denoted by \(\omega\) and \(\omega^2\). Let's assign \(\alpha\) and \(\beta\) to these values. Without loss of generality, let \(\alpha = \omega = \frac{{ - 1 + i\sqrt 3 }}{2}\) and \(\beta = \omega^2 = \frac{{ - 1 - i\sqrt 3 }}{2}\).

Properties of Complex Cube Roots of Unity

The complex cube roots of unity (\(1, \omega, \omega^2\)) have important properties:

• The sum of the three cube roots of unity is zero: \(1 + \omega + \omega^2 = 0\). This implies \(\omega + \omega^2 = -1\).
• The cube of \(\omega\) is 1: \(\omega^3 = 1\).
• Higher powers of \(\omega\) repeat in a cycle of 3: \(\omega^4 = \omega^3 \cdot \omega = 1 \cdot \omega = \omega\), \(\omega^5 = \omega^3 \cdot \omega^2 = 1 \cdot \omega^2 = \omega^2\), \(\omega^6 = (\omega^3)^2 = 1^2 = 1\), and so on. In general, \(\omega^{3k} = 1\), \(\omega^{3k+1} = \omega\), and \(\omega^{3k+2} = \omega^2\) for any integer \(k\).
• Similarly, powers of \(\omega^2\) follow a similar pattern because \((\omega^2)^j = \omega^{2j}\). \((\omega^2)^3 = \omega^6 = (\omega^3)^2 = 1^2 = 1\).

Evaluating the Summation

The summation we need to evaluate is \(\mathop \sum \limits_{j = 0}^3 \left( {{\alpha ^j} + {\beta ^j}} \right)\). This sum expands to:

\(\left( {{\alpha ^0} + {\beta ^0}} \right) + \left( {{\alpha ^1} + {\beta ^1}} \right) + \left( {{\alpha ^2} + {\beta ^2}} \right) + \left( {{\alpha ^3} + {\beta ^3}} \right)\)

Let's calculate each term separately, using \(\alpha = \omega\) and \(\beta = \omega^2\):

• For \(j=0\): \({\alpha ^0} + {\beta ^0} = \omega^0 + (\omega^2)^0 = 1 + 1 = 2\). (Any non-zero number raised to the power of 0 is 1).
• For \(j=1\): \({\alpha ^1} + {\beta ^1} = \omega^1 + (\omega^2)^1 = \omega + \omega^2\). Using the property \(1 + \omega + \omega^2 = 0\), we get \(\omega + \omega^2 = -1\).
• For \(j=2\): \({\alpha ^2} + {\beta ^2} = \omega^2 + (\omega^2)^2 = \omega^2 + \omega^4\). Since \(\omega^4 = \omega^3 \cdot \omega = 1 \cdot \omega = \omega\), this term becomes \(\omega^2 + \omega\). Again, using the property \(1 + \omega + \omega^2 = 0\), we get \(\omega^2 + \omega = -1\).
• For \(j=3\): \({\alpha ^3} + {\beta ^3} = \omega^3 + (\omega^2)^3\). Using the property \(\omega^3 = 1\) and \((\omega^2)^3 = \omega^6 = (\omega^3)^2 = 1^2 = 1\), this term becomes \(1 + 1 = 2\).

Calculating the Total Sum

Now, we sum the results for each value of \(j\):

Total Sum \( = (\alpha^0 + \beta^0) + (\alpha^1 + \beta^1) + (\alpha^2 + \beta^2) + (\alpha^3 + \beta^3) \)

Total Sum \( = 2 + (-1) + (-1) + 2 \)

Total Sum \( = 2 - 1 - 1 + 2 \)

Total Sum \( = 0 + 2 = 2 \)

Therefore, the value of the summation \(\mathop \sum \limits_{j = 0}^3 \left( {{\alpha ^j} + {\beta ^j}} \right)\) is 2.

Adding the values from the last column: \(2 + (-1) + (-1) + 2 = 2\).

Revision Table: Quadratic Equation Roots

Additional Information: Complex Cube Roots of Unity

The complex cube roots of unity are the solutions to the equation \(x^3 = 1\). These roots are \(1\), \(\omega\), and \(\omega^2\). They can be represented in polar form:

• \(1 = \cos(0) + i\sin(0)\)
• \(\omega = e^{i 2\pi/3} = \cos(2\pi/3) + i\sin(2\pi/3) = -\frac{1}{2} + i\frac{\sqrt{3}}{2}\)
• \(\omega^2 = e^{i 4\pi/3} = \cos(4\pi/3) + i\sin(4\pi/3) = -\frac{1}{2} - i\frac{\sqrt{3}}{2}\)

Notice that \(x^2 + x + 1\) is a factor of \(x^3 - 1\), since \(x^3 - 1 = (x-1)(x^2+x+1)\). The roots of \(x^2+x+1=0\) are therefore the roots of \(x^3-1=0\) other than \(x=1\), which are \(\omega\) and \(\omega^2\).