Question

$if \: x = ( \sqrt{2} + 1) {}^{ - \frac{1}{3} \: } then \: the \: \\ value \: of \: ( {x}^{3} + \frac{1}{ {x}^{3} } ) \: is$

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Answer 1

Answer:

f(x) = kx³ – 8x² + 5

Roots are α – β , α & α +β

Sum of roots = – (-8)/k

Sum of roots = α – β + α + α +β = 3α

= 3α = 8/k

= k = 8/3α

or we can solve as below

f(x) = (x – (α – β)(x – α)(x – (α +β))

= (x – α)(x² – x(α+β + α – β) + (α² – β²))

= (x – α)(x² – 2xα + (α² – β²))

= x³ – 2x²α + x(α² – β²) – αx² +2α²x – α³ + αβ²

= x³ – 3αx² + x(3α² – β²) + αβ² – α³

= kx³ – 3αkx² + xk(3α² – β²) + k(αβ² – α³)

comparing with

kx³ – 8x² + 5

k(3α² – β²) = 0 => 3α² = β²

k(αβ² – α³) = 5

=k(3α³ – α³) = 5

= k2α³ = 5

3αk = 8 => k = 8/3α

(8/3α)2α³ = 5

=> α² = 15/16

=> α = √15 / 4

Answer 2

Answer:

I am not an expert, I am an ace :P.

GIVEN:

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

WE NEED TO FIND THE VALUE OF:

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

Now,

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

Cube on both sides.

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

Look at the RHS. we know that :

(xᵃ)ᵇ = xᵃᵇ.

So, $x^{3 } =((\sqrt{2} +1)^{-\frac{1}{3} })^3$ can be written as:

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

$x^{3 } =((\sqrt{2} +1)^{-1})$

Look at RHS. we know that:

x⁻ⁿ = 1/xⁿ.

So, $x^{3 } =((\sqrt{2} +1)^{-1})$ can be written as:

$x^{3 } =\frac{1}{\sqrt{2} +1}$.

Now, substitute the value of x³ in $x^{3} +\frac{1}{x^{3} }.$

$\to \frac{1}{\sqrt{2} +1}+\frac{1}{\frac{1}{\sqrt{2}+1 } }$

$\to \frac{1}{\sqrt{2} +1}+{\sqrt{2}+1 }$

Rationalise $\frac{1}{\sqrt{2} +1}$.

$\to \frac{1 \times \sqrt{2} -1}{(\sqrt{2} +1)\times(\sqrt{2} -1)}+{\sqrt{2}+1 }$

$\to{\sqrt{2} -1}+\sqrt{2} +1$

$\to 2\sqrt{2}$

$\therefore x^{3} +\frac{1}{x^{3} }=2\sqrt{2}$

HOPE THIS HELPS :D

#Quality.