Simplify in the form: $x^n + \frac{1}{x^n}$

Question

Simplify: $$\left(x^2 + \sqrt2 + \frac{1}{x^2}\right)\left(x^2 - \sqrt2 + \frac{1}{x^2}\right)$$ in the form $x^n + \frac{1}{x^n}$

Answer 1

$$\begin{align} & \left(x^2 + \sqrt2 + \frac{1}{x^2}\right)\left(x^2 - \sqrt2 + \frac{1}{x^2}\right) \\ & =\left(x^2 + \frac{1}{x^2}\right)^2 - \left(\sqrt2\right)^2 \\ & =\left(x^2 + \frac{1}{x^2}\right)^2 - 2 \\ & =x^4 + \frac{1}{x^4} + 2 - 2 \\ & =x^4 + \frac{1}{x^4} \end{align}$$

Hope this helps.

Answer 2

$$\left(x^2 + \sqrt2 + \frac{1}{x^2}\right)\left(x^2 - \sqrt2 + \frac{1}{x^2}\right)$$

=$$\left(x^2 + \frac{1}{x^2}+\sqrt2\right)\left(x^2 + \frac{1}{x^2}-\sqrt2\right)$$

What can you say about the form:

$$(a-b)(a+b)$$

Answer 3

One may write $$ \begin{align} \left(x^2 + \sqrt2 + \frac{1}{x^2}\right)\left(x^2 - \sqrt2 + \frac{1}{x^2}\right)&=\left(x^2 + \frac{1}{x^2}+ \sqrt2\right)\left(x^2 + \frac{1}{x^2} - \sqrt2\right) \\&=\left[\left(x^2 + \frac{1}{x^2}\right)^2- 2\right] \\&=x^4+2 + \frac{1}{x^4}- 2 \\&=x^4 + \frac{1}{x^4} \end{align} $$

Answer 4

$$\begin{align} & \left(x^2 + \frac{1}{x^2}\right)^2 - \left(\sqrt2\right)^2 \\ & =x^4 + \frac{1}{x^4} +2 \cdot \frac{1}{x^2} \cdot x^2 -\left(\sqrt2\right)^2 \\ & =x^4 + \frac{1}{x^4} +2 -2 \\ & =x^4 + \frac{1}{x^4}. \end{align}$$