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If $x+\dfrac{1}{x} = 4$, then what is the value of $x^6 + \dfrac{1}{x^6} ?$

1. $52$
2. $256$
3. $1026$
4. $2702$
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Answer will be $2702$

$x+\dfrac{1}{x} = 4$

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, $\color{Green}{x^{6}+\dfrac{1}{x^{6}}} = 2744-42 = \color{pink}{2702}$
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