Question

x+1/x=3 then find the value of x^6+1/x^6 =?

Answer 1

$x + \frac{1}{x} = 3 \\ \\ \\ (x + \frac{1}{x})^2 = 3^2 \\ \\ = x^2 + \frac{1}{x^2} + (2 \times x \times \frac{1}{x}) = 9 \\ \\ = x^2 + \frac{1}{x^2} + 2 = 9 \\ \\ x^2 + \frac{1}{x^2} = 9 - 2 = 7 \\ \\ \\ (x^2 + \frac{1}{x^2})^2 = 7^2 \\ \\ = x^4 + \frac{1}{x^4} + (2 \times x^2 \times \frac{1}{x^2}) = 49 \\ \\ = x^4 + \frac{1}{x^4} + 2 = 49 \\ \\ x^4 + \frac{1}{x^4} = 49 - 2 = 47$

$\\ \\ \\ (x + \frac{1}{x})^6 = 3^6 \\ \\ = x^6 + (6 \times x^5 \times \frac{1}{x}) + (15 \times x^4 \times \frac{1}{x^2}) + (20 \times x^3 \times \frac{1}{x^3}) + (15 \times x^2 \times \frac{1}{x^4}) + (6 \times x \times \frac{1}{x^5}) + \frac{1}{x^6} = 729 \\ \\ = x^6 + 6x^4 + 15x^2 + 20 + \frac{15}{x^2} + \frac{6}{x^4} + \frac{1}{x^6} = 729 \\ \\ = x^6 + 6(x^4 + \frac{1}{x^4}) + 15(x^2 + \frac{1}{x^2}) + 20 + \frac{1}{x^6} = 729 \\ \\ = x^6 + (6 \times 47) + (15 \times 7) + 20 + \frac{1}{x^6} = 729$

$\\ \\ \\ = x^6 + 282 + 105 + 20 + \frac{1}{x^6} = 729 \\ \\ = x^6 + \frac{1}{x^6} + 407 = 729 \\ \\ \\ x^6 + \frac{1}{x^6} = 729 - 407 \\ \\ = x^6 + \frac{1}{x^6} = 322$

∴ 322 is the answer.

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Answer 2

Answer:

The value of $x^{6} +\frac{1}{x^{6} }$ is 322.

Step-by-step explanation:

Given an equation, $x+\frac{1}{x} =3$

cubing on both sides,

                               $\big(x+\frac{1}{x}\big)^{3} =3^{3}$

 $x^{3} +\frac{1}{x^{3} }+3\times x\times\frac{1}{x} \big(x+\frac{1}{x}\big)=27$

              $x^{3} +\frac{1}{x^{3} }+3\big(x+\frac{1}{x}\big)=27$

                     $x^{3} +\frac{1}{x^{3} }+3(3)=27$

                                $x^{3} +\frac{1}{x^{3} }=18$

Given to find $x^{6} +\frac{1}{x^{6} }$, which can be written as

              $(x^{3})^{2} +\frac{1}{(x^{3})^{2} }=\big(x^{3}+\frac{1}{x^{3}} }\big)^{2} -2\times x \times \frac{1}{x}$

                                  $=(18)^{2} -2=324-2=322$

Therefore, the value of $x^{6} +\frac{1}{x^{6} }$ is 322.