Question

If, x+1/x = √3 ,Find,x⁶+1

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Answer 1

Given,

$\longrightarrow x+\dfrac{1}{x}=\sqrt3$

$\longrightarrow\dfrac{x^2+1}{x}=\sqrt3$

$\longrightarrow x^2+1=x\sqrt3$

Squaring both sides,

$\longrightarrow (x^2+1)^2=(x\sqrt3)^2$

$\longrightarrow x^4+2x^2+1=3x^2$

$\longrightarrow x^4-x^2+1=0\quad\quad\dots(1)$

Multiplying both sides of (1) by $x^2,$

$\longrightarrow x^6-x^4+x^2=0\quad\quad\dots(2)$

Now, adding (1) and (2),

$\longrightarrow x^6-x^4+x^2+x^4-x^2+1=0+0$

$\longrightarrow\underline{\underline{x^6+1=0}}$

Hence 0 is the answer.

Answer 2

Answer:

⟶x+

x

1

=

3

\longrightarrow\dfrac{x^2+1}{x}=\sqrt3⟶

x

x

2

+1

=

3

\longrightarrow x^2+1=x\sqrt3⟶x

2

+1=x

3

Squaring both sides,

\longrightarrow (x^2+1)^2=(x\sqrt3)^2⟶(x

2

+1)

2

=(x

3

)

2

\longrightarrow x^4+2x^2+1=3x^2⟶x

4

+2x

2

+1=3x

2

\longrightarrow x^4-x^2+1=0\quad\quad\dots(1)⟶x

4

−x

2

+1=0…(1)

Multiplying both sides of (1) by x^2,x

2

,

\longrightarrow x^6-x^4+x^2=0\quad\quad\dots(2)⟶x

6

−x

4

+x

2

=0…(2)

Now, adding (1) and (2),

\longrightarrow x^6-x^4+x^2+x^4-x^2+1=0+0⟶x

6

−x

4

+x

2

+x

4

−x

2

+1=0+0

\longrightarrow\underline{\underline{x^6+1=0}}⟶

x

6

+1=0

Hence 0 is the answer.