Question

$if \: {x}^{2} - 3x + 1 = 0 \: then \: value \: of \: {x}^{6} + \frac{1}{ {x}^{6} }$


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Answer 1

$x^2-3x+1=0\\\\\\x^2+1=3x\\\\\\x+\dfrac{1}{x}=3\quad\longrightarrow\quad(1)$

From (1),

$\left(x+\dfrac{1}{x}\right)^2=3^2\\\\\\x^2+\dfrac{1}{x^2}+2=9\\\\\\x^2+\dfrac{1}{x^2}=7\quad\longrightarrow\quad(2)$

Again from (1),

$\left(x+\dfrac{1}{x}\right)^3=3^3\\\\\\x^3+\dfrac{1}{x^3}+3\left(x+\dfrac{1}{x}\right)=27\\\\\\x^3+\dfrac{1}{x^3}+3\times3=27\\\\\\x^3+\dfrac{1}{x^3}+9=27\\\\\\x^3+\dfrac{1}{x^3}=18\quad\longrightarrow\quad(3)$

Finally, multiplying (2) and (3),

$\left(x^2+\dfrac{1}{x^2}\right)\left(x^3+\dfrac{1}{x^3}\right)=7\times18\\\\\\x^6+\dfrac{1}{x^6}+x+\dfrac{1}{x}=126\\\\\\x^6+\dfrac{1}{x^6}+3=126\quad[\text{From (1)}]\\\\\\x^6+\dfrac{1}{x^6}=\mathbf{123}$