Factorise: ${{x}^{2}}+\dfrac{1}{{{x}^{2}}}-3$.

Hint: We first try to explain the concept of factorisation and the ways a factorisation of a polynomial can be done. We use the identity theorem of ${{a}^{2}}-{{b}^{2}}=\left( a+b \right)\left( a-b \right)$ to factor the given polynomial ${{x}^{2}}+\dfrac{1}{{{x}^{2}}}-3$. We assume the values of $a=\left( x-\dfrac{1}{x} \right);b=1$. The final multiplied linear polynomials are the solution of the problem.

Complete step-by-step answer:
The main condition of factorisation is to break the given number or function or polynomial into multiple of basic primary numbers or polynomials.
The simplification of ${{x}^{2}}+\dfrac{1}{{{x}^{2}}}-3$ gives
\[\begin{align}
  & {{x}^{2}}+\dfrac{1}{{{x}^{2}}}-3 \\
 & ={{x}^{2}}+\dfrac{1}{{{x}^{2}}}-2\times x\times \dfrac{1}{x}-1 \\
 & ={{\left( x-\dfrac{1}{x} \right)}^{2}}-{{1}^{2}} \\
\end{align}\]
For the factorisation of the given quadratic polynomial ${{x}^{2}}+\dfrac{1}{{{x}^{2}}}-3$, we apply the factorisation identity of difference of two squares as ${{a}^{2}}-{{b}^{2}}=\left( a+b \right)\left( a-b \right)$.
We get \[{{x}^{2}}+\dfrac{1}{{{x}^{2}}}-3={{\left( x-\dfrac{1}{x} \right)}^{2}}-{{1}^{2}}\]. We put the value of $a=\left( x-\dfrac{1}{x} \right);b=1$.
Factorisation of the polynomial gives us
${{\left( x-\dfrac{1}{x} \right)}^{2}}-{{1}^{2}}=\left( x-\dfrac{1}{x}+1 \right)\left( x-\dfrac{1}{x}-1 \right)$.

Note: The highest power of the variable or the degree of a polynomial decides the number of roots or the solution of that polynomial. Quadratic equations have two roots. Cubic polynomials have three. It can have both real and imaginary roots. In the given polynomial the values will be dependent on the second variable.