Factorise the following:

$\dfrac{x^6}{343} + \dfrac{343}{x^6}$.

$\dfrac{x^6}{343} + \dfrac{343}{x^6} = \Big(\dfrac{x^2}{7}\Big)^3 + \Big(\dfrac{7}{x^2}\Big)^3$.

We know that,

a³ + b³ = (a + b)(a² - ab + b²)

$\therefore \Big(\dfrac{x^2}{7}\Big)^3 + \Big(\dfrac{7}{x^2}\Big)^3 = \Big(\dfrac{x^2}{7} + \dfrac{7}{x^2}\Big)\Big[\Big(\dfrac{x^2}{7}\Big)^2 - \dfrac{x^2}{7} \times \dfrac{7}{x^2} + \Big(\dfrac{7}{x^2}\Big)^2\Big] \\[1em] = \Big(\dfrac{x^2}{7} + \dfrac{7}{x^2}\Big)\Big(\dfrac{x^4}{49} - 1 + \dfrac{49}{x^4}\Big).$

Hence, $\dfrac{x^6}{343} + \dfrac{343}{x^6} = \Big(\dfrac{x^2}{7} + \dfrac{7}{x^2}\Big)\Big(\dfrac{x^4}{49} - 1 + \dfrac{49}{x^4}\Big).$