Square it, then cube it

Let $A=x$ and $B= \frac{1}{x},$ . We have that $A+B=3$ and $AB=1$ . Therefore, $(A + B)^2 = 9$ or equivalently, $A^2 + 2AB + B^2 = 9$ , which implies $A^2+B^2=7$ .

$x^6+ \frac{1}{x^6} = A^6+ B^6 \\ =(A^2+B^2) ({(A^2+B^2)}^2-3{(AB)}^2 ) \\= {(7) (7^2-3{(1)}^2}) \\= \boxed{322}$

$x^6 + \frac{1}{x^6} = ? \\ \Rightarrow \big(x^3\big)^2 + \bigg(\frac{1}{x^3}\bigg)^2 + 2 - 2 = ? \\ \Rightarrow \bigg(x^3 + \frac{1}{x^3}\bigg)^2 - 2 = ? \\ \Rightarrow \Bigg(\bigg(x + \frac{1}{x}\bigg)\bigg(x^2 - 1 + \frac{1}{x^2}\bigg)\Bigg)^2 - 2 = ? \\ \Rightarrow \Bigg(\bigg(x + \frac{1}{x}\bigg)\bigg(x^2 - 1 + \frac{1}{x^2} + 2 - 2\bigg)\Bigg)^2 - 2 = ? \\ \Rightarrow \Bigg(\bigg(x + \frac{1}{x}\bigg)\bigg(\Big(x + \frac{1}{x}\Big)^2 - 3\bigg)\Bigg)^2 - 2 = ? \\ \large{\stackrel{x + \frac{1}{x} = 3}{\Rightarrow}} \Bigg(3\cdot\Big(3^2 - 3 \Big) \Bigg)^2 - 2 = ? \\ \large{\Rightarrow \boxed{? = 322}}$

x+1/x = 3. Then (x+1/x)^3 = 3^3 = 27 ie x^3 + 1/x^3 + 3x.1/x(x + 1/x) = 27 ie x^3+1/x^3 +3x3 = 27 Hence x^3 + 1/x^3 = 27-9

x^3 + 1/x^3 =18 So (x^3 + 1/x^3)^2 = 18^2, Then x^6 + 1/x^6 + 2x.1/x = 324 , Therefore x^6 + 1/x^6 = 324 - 2 =322

x^2+1/x^2=3
squaring both sides,
x^2+1/x^2=7
now cubing both sides.​
x^6+1/x^6+3(x^2+1/x^2)=343
x^6+1/x^6+3*7=343
x^6+1/x^6=343-21=322.

since (x +1/x)^2= x^2+ 1/x^2 + 2 Therefore x^2 +1/x^2 = 9-2 =7 Again take the cube of this We get [x^2+1/x^2]^3 = x^6 + 1/x^6+3(7) Solving it we get x^6+1/x^6 = 7^3 -21 finally we get 343-21 That is 322

you dont need to square anything just cube first function and make perfect square of second term

x+1/x=3 so squaring both sides (x+1/x)^2=3^2 which is x^2+1/x^2+2(x 1/x)=9 then x^2+1/x^2+2=9 then x^2+1/x^2=7 then cubing both sides (x^2+1/x^2)^3=7^3 which is x^6+1/x^6+3(x^2 1/x^2)(x^2+1/x^2)=343 then x^6+1/x^6+3(7)=343 {i.e. x^2+1/x^2)=7} then x^6+1/x^6=322

x+1/x= a then x^2+1/x^2= (x+1/x)^2-2. -(1)

x^3+1/x^3 =(x+1/x)^3-3(x+1/x) (Since (a+b)^3=a^3+b^3+3ab(a+b)) -(2)

Using (2) x^3+1/x^3=27-3(3)=18

Using (1) x^6+1/x^6= 18^2-2=322