this is for 8th grade pls answer .

Step-by-step explanation:
We have that
[tex](x + \frac{1}{x} ) {}^{2} = 3[/tex]
We are trying to find the number value so that we can apply in the later equation.
Qe first simplify.
Remeber that
[tex](a + b) {}^{2} = a {}^{2} + 2ab + {b}^{2} [/tex]
Also remeber that
[tex] \frac{1}{x} = {x}^{ - 1} [/tex]
so
[tex](x + x {}^{ - 1} ) {}^{2} = {x}^{2} + 2x {}^{0} + {x}^{ - 2} = 3[/tex]
We then simply remeber that x^0=1 so
[tex] {x}^{2} + 2 + \frac{1}{ {x}^{2} } = 3[/tex]
Multiply both sides by x^2.
[tex] {x}^{4} + 2 {x}^{2} + 1 = 3 {x}^{2} [/tex]
Subtract both sides by 3x^2
[tex] {x}^{4} - {x}^{2} + 1 = 0[/tex]
Notice that x^4= (x^2)^2.
So our reformed equation is
[tex]( {x}^{2} ) {}^{2} - {x}^{2} + 1 = 0[/tex]
Let a variable , w equal x^2. This means that we subsitute variable, w in for x^2.
[tex]w {}^{2} - w + 1 = 0[/tex]
Now we use the quadratic formula
[tex] w = \frac{ - b + \sqrt{b {}^{2} - 4ac } }{2a} [/tex]
and
[tex]w = - b - \frac { \sqrt{b {}^{2} - 4ac } }{2a} [/tex]
Let a=1 b=-1 and c=1.
[tex]w = \frac{1 + \sqrt{1 - 4(1)} }{2} [/tex]
[tex]w = \frac{1 - \sqrt{1 - 4} }{2} [/tex]
Now, we get
[tex]w = \frac{1}{2} + \frac{i \sqrt{3} }{2} [/tex]
and
[tex]w = \frac{1}{2} - \frac{ i\sqrt{3} }{2} [/tex]
Now since we set both of these to the x^2 we solve for x.
and
[tex] {x}^{2} = \frac{1}{2} + \frac{i \sqrt{3} }{2} [/tex]
and
[tex] {x}^{2} = \frac{1}{2} - \frac{i \sqrt{3} }{2} [/tex]
We can represent both of these as complex number in the form of a+bi. Next we can convert this into trig form which is
[tex] {x}^{2} = 1( \cos(60) + i \: \sin(60) [/tex]
and
[tex] {x}^{2} = 1( \cos(300) + i \: sin(300))[/tex]
Next we take the sqr root of 1 which is 1, and divide the degree by two.
[tex] {x} = 1( \cos(30) + i \: sin \: 30)[/tex]
and
[tex]x = 1( \cos(150) + i \: sin(150)[/tex]
We are asked for the 2nd root so just add 180 degrees to this and we have
[tex]x = 1 \cos(210) + i \: sin \: 210)[/tex]
and
[tex]x = 1( \cos(330) + i \: sin(330)[/tex]
which both simplified to
[tex]x = - \frac{ \sqrt{3} }{2} - \frac{1}{2} i[/tex]
and
[tex]x = \frac{ \sqrt{3} }{2} - \frac{1}{2} i[/tex]
Now we must find
x^18+x^12+x^6+1.
We just use demovire Theorem. Which is a complex number raised to the nth root is
[tex] {r}^{n} (cos(nx) + i \: sin(nx)[/tex]
So let plug in our first root.
[tex]1( \cos(330 \times 18)) + i \: sin \: (330 \times 18))) + 1( \cos(12 \times 330)) + i \: sin(12 \times 330) + 1( \cos(6 \times 330) + i \: sin(6 \times 330))) + 1[/tex]
To save time we multiply the angle and use rules of terminals angle and we get
[tex]1( \cos(180) + i \sin(180) ) + 1( \cos(0) + i \: sin \:( 0) + 1( \cos(180) + i \: sin(180) + 1[/tex]
And we get
[tex] - 1 + 1 + - 1 + 1 = 0[/tex]
So one of the answer is x=0
And the other, let see
[tex]1 \cos(210 \times 18) + i \: \sin(210 \times 18) + 1 \: cos(210 \times 12) + i \: sin(210 \times 12) + 1 \cos(210 \times 6) + \:i sin(210 \times 6) + 1[/tex]
[tex] \cos(180) + i \: sin(180) + 1 \cos(0) + i\sin(0) +1( \cos(0) + i \sin(0) + 1[/tex]
We get
[tex] - 1 + 1 + 1 + 1 = 2[/tex]
So our answer are 2.
So the answer to the second part is
0 and 2.