Solving the Quadratic Equation for \(x^6 + \frac{1}{x^6}\)

We are given the quadratic equation \(x^2 + 3x + 1 = 0\) and asked to find the value of \({x^6}\; + \;\frac{1}{{{x^6}}}.\) This problem can be solved by first finding the value of \(x + \frac{1}{x}\) and then using algebraic identities to find the values of \(x^2 + \frac{1}{x^2}\) and finally \(x^6 + \frac{1}{x^6}\).

Step-by-Step Calculation for \(x^6 + \frac{1}{x^6}\)

Let's start with the given equation:

$$x^2 + 3x + 1 = 0$$

Since the coefficients are non-zero, \(x\) cannot be zero. We can divide the entire equation by \(x\) to get a relationship between \(x\) and \(\frac{1}{x}\).

Divide by \(x\):

$$\frac{x^2}{x} + \frac{3x}{x} + \frac{1}{x} = \frac{0}{x}$$

This simplifies to:

$$x + 3 + \frac{1}{x} = 0$$

Rearranging the terms, we get the value of \(x + \frac{1}{x}\):

$$x + \frac{1}{x} = -3$$

Now that we have \(x + \frac{1}{x}\), we can find \(x^2 + \frac{1}{x^2}\) by squaring both sides of the equation \(x + \frac{1}{x} = -3\).

Square both sides:

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

Using the identity \((a+b)^2 = a^2 + 2ab + b^2\), where \(a=x\) and \(b=\frac{1}{x}\):

$$x^2 + 2\left(x\right)\left(\frac{1}{x}\right) + \left(\frac{1}{x}\right)^2 = 9$$

$$x^2 + 2 + \frac{1}{x^2} = 9$$

Subtract 2 from both sides to isolate \(x^2 + \frac{1}{x^2}\):

$$x^2 + \frac{1}{x^2} = 9 - 2$$

$$x^2 + \frac{1}{x^2} = 7$$

Next, we need to find \(x^6 + \frac{1}{x^6}\). We can obtain \(x^6\) by cubing \(x^2\). So, we can cube the expression \(x^2 + \frac{1}{x^2}\) that we just found. We use the identity \(a^3 + b^3 = (a+b)^3 - 3ab(a+b)\), where \(a=x^2\) and \(b=\frac{1}{x^2}\).

We want to find \( (x^2)^3 + \left(\frac{1}{x^2}\right)^3 \), which is \(x^6 + \frac{1}{x^6}\).

Using the identity:

$$x^6 + \frac{1}{x^6} = \left(x^2 + \frac{1}{x^2}\right)^3 - 3\left(x^2\right)\left(\frac{1}{x^2}\right)\left(x^2 + \frac{1}{x^2}\right)$$

$$x^6 + \frac{1}{x^6} = \left(x^2 + \frac{1}{x^2}\right)^3 - 3\left(x^2 + \frac{1}{x^2}\right)$$

Substitute the value \(x^2 + \frac{1}{x^2} = 7\) into this equation:

$$x^6 + \frac{1}{x^6} = (7)^3 - 3(7)$$

Calculate the values:

$$7^3 = 7 \times 7 \times 7 = 49 \times 7 = 343$$

$$3 \times 7 = 21$$

Substitute these values back:

$$x^6 + \frac{1}{x^6} = 343 - 21$$

$$x^6 + \frac{1}{x^6} = 322$$

Thus, the value of \({x^6}\; + \;\frac{1}{{{x^6}}}\) is 322.

Summary of Intermediate Values

Revision Table: Key Algebraic Identities for Solving Power Problems

Additional Information on Solving Equations and Algebraic Identities

The technique used here is common for problems involving expressions of the form \(x^n + \frac{1}{x^n}\) or \(x^n - \frac{1}{x^n}\) when a relationship like \(x \pm \frac{1}{x} = k\) is known or can be derived from a given equation (like a quadratic equation).

For a quadratic equation \(ax^2 + bx + c = 0\), if \(a=c\), dividing by \(x\) often leads to a simple form of \(x + \frac{1}{x} = -\frac{b}{a}\). In our given equation \(x^2 + 3x + 1 = 0\), \(a=1\) and \(c=1\), so \(a=c\). Dividing by \(x\) gives \(x + 3 + \frac{1}{x} = 0\), which results in \(x + \frac{1}{x} = -3\), confirming this pattern.

Once \(x \pm \frac{1}{x}\) is found, you can iteratively find higher powers:

• \(x^2 + \frac{1}{x^2}\) from \((x + \frac{1}{x})^2\) or \((x - \frac{1}{x})^2\).
• \(x^3 + \frac{1}{x^3}\) from \((x + \frac{1}{x})^3\).
• \(x^4 + \frac{1}{x^4}\) from \((x^2 + \frac{1}{x^2})^2\).
• \(x^5 + \frac{1}{x^5}\) from \((x^2 + \frac{1}{x^2})(x^3 + \frac{1}{x^3}) - (x + \frac{1}{x})\).
• \(x^6 + \frac{1}{x^6}\) from \((x^2 + \frac{1}{x^2})^3\) or \((x^3 + \frac{1}{x^3})^2\).

In this specific problem, to get \(x^6 + \frac{1}{x^6}\), we cubed \(x^2 + \frac{1}{x^2}\). Alternatively, we could have first calculated \(x^3 + \frac{1}{x^3}\) from \(x + \frac{1}{x}\), and then squared the result to get \(x^6 + \frac{1}{x^6}\). Both methods yield the same result:

• Method 1: Find \(x^2 + \frac{1}{x^2}\), then cube it to find \(x^6 + \frac{1}{x^6}\).
• Method 2: Find \(x^3 + \frac{1}{x^3}\), then square it to find \(x^6 + \frac{1}{x^6}\).

Both methods rely heavily on the fundamental algebraic identities relating sums and products of variables and their reciprocals.