If $x - \dfrac{1}{x} = 5$, find the value of $x^4 + \dfrac{1}{x^4}$.

$x^4 + \dfrac{1}{x^4} = (x^2)^2 + \Big(\dfrac{1}{x^2}\Big)^2 …..[i] \\[1em] \Big(x^2 + \dfrac{1}{x^2}\Big)^2 = (x^2)^2 + 2 \times (x^2)^2 \times \dfrac{1}{x^2} + \Big(\dfrac{1}{x^2}\Big)^2 \\[1em] \therefore (x^2)^2 + \Big(\dfrac{1}{x^2}\Big)^2 = \Big(x^2 + \dfrac{1}{x^2}\Big)^2 - 2 \\[1em]$

Putting this value of $(x^2)^2 + \Big(\dfrac{1}{x^2}\Big)^2$ in eqn (i), we get:

$x^4 + \dfrac{1}{x^4} = \Big(x^2 + \dfrac{1}{x^2}\Big)^2 - 2 \space …..[ii] \\[1em] \Big(x - \dfrac{1}{x}\Big)^2 = x^2 - 2 \times x \times \dfrac{1}{x} + \Big(\dfrac{1}{x}\Big)^2 \\[1em] \therefore x^2 + \dfrac{1}{x^2} = \Big(x - \dfrac{1}{x}\Big)^2 + 2 \\[1em]$

Putting this value of $x^2 + \dfrac{1}{x^2}$ in eqn (ii), we get:

$x^4 + \dfrac{1}{x^4} = \Big[\Big(x - \dfrac{1}{x}\Big)^2 + 2\Big]^2 - 2 \\[1em]$

Putting the given values in above eqn, we get:

$x^4 + \dfrac{1}{x^4} = (5^2 + 2)^2 - 2 \\[1em] = (27)^2 - 2 \\[1em] = 729 - 2 \\[1em] = 727.$

Hence, $x^4 + \dfrac{1}{x^4}$ = 727.