$x+\dfrac{1}{x} = 4$
squaring on both sides , we get -
Or, $\left (x+\dfrac{1}{x} \right)^2 = 4^2$
Or, $x^2+\left (\dfrac{1}{x} \right )^2 + 2.x.\dfrac{1}{x} = 16$
Or, $x^2+\left (\dfrac{1}{x} \right )^2 + 2 = 16$
Or, $x^2+\dfrac{1}{x^2}= 16-2 = 14$ ------------------1)
Now, we'll cubing on both sides-
$\left (x^2+\dfrac{1}{x^2} \right)^3 = (14)^3$
Or, $ (x^2)^3+\left(\dfrac{1}{x^2}\right)^3 + 3.(x^2)^2.\dfrac{1}{x^2}+3.x^2.\left(\dfrac{1}{x^2}\right)^2 = (14)^3 = 2744$ $ \color{purple}{\left [\because (a+b)^3 = a^3+b^3+3a^2b+3ab^2 \right ]}$
Or, $ x^{2 \times 3}+\left(\dfrac{1}{x^{2 \times 3}}\right) + 3.x^{2 \times 2}.\dfrac{1}{x^2}+3.x^2.\left(\dfrac{1}{x^{2 \times 2}}\right) =2744$
Or, $x^{6}+\dfrac{1}{x^{6}} + 3.x^{4}.\dfrac{1}{x^2}+3.x^2.\dfrac{1}{x^{4}}=2744$
Or, $x^{6}+\dfrac{1}{x^{6}} + 3.x^{2}+3.\dfrac{1}{x^{2}}=2744$
Or, $x^{6}+\dfrac{1}{x^{6}} + 3\left (x^{2}+\dfrac{1}{x^{2}} \right )=2744$
Or, $x^{6}+\dfrac{1}{x^{6}} + 3\left (14 \right )=2744$ $ \qquad \qquad \left [ \color{Blue }{ \because \text{We know,} \left (x^{2}+\dfrac{1}{x^{2}} \right ) = 14} \color{teal}{\text{ ----from 1)} }\right ]$
Or, $x^{6}+\dfrac{1}{x^{6}} + 42 = 2744$
Or, $\color{Green}{x^{6}+\dfrac{1}{x^{6}}} = 2744-42 = \color{pink}{2702}$