Question

if x =1+√2 then show that (x-1/x)^3 =8​

Answer 1

$(x-\dfrac{1}{x})^3$ = 8​, proved.

Step-by-step explanation:

We have,

x = 1 + $\sqrt{2}$

Show that, $(x-\dfrac{1}{x})^3$ = 8​.

∴ $\dfrac{1}{x}=\dfrac{1}{1+\sqrt{2}}$

⇒ $\dfrac{1}{x}=\dfrac{1}{\sqrt{2}+1}$

Rationalising numerator and denominator, we get

$\dfrac{1}{x}=\dfrac{1}{\sqrt{2}+1}\times \dfrac{\sqrt{2}-1}{\sqrt{2}-1}$

⇒ $\dfrac{1}{x}=\dfrac{\sqrt{2}-1}{\sqrt{2}^2-1^2}=\dfrac{\sqrt{2}-1}{2-1}$

⇒ $\dfrac{1}{x}$ = $\sqrt{2}$ - 1

L.H.S. = $(x-\dfrac{1}{x})^3$

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

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

=$(1+1)^3$

= $(2)^3$

= 8 = R.H.S., shown.

Answer 2

Answer:

Given,

x=1+

2

​

.

Now,

(x−

x

1

​

)

3

=(1+

2

​

−(

1+

2

​

1

​

))

3

=[1+

2

​

−(

2

​

+1

1

​

)×(

2

​

−1

2

​

−1

​

)]

3

=[1+

2

​

−(

2−1

2

​

−1

​

)]

3

=[1+

2

​

−(

2

​

−1)]

3

=(1+

2

​

−

2

​

+1)

3

=2

3

=8.

Step-by-step explanation: