Question

if x = 2 - $\sqrt{3}$, then find the value of $x^{3}$ - $\frac{1}{x^{3}}$

Answer 1

Answer:

please type something sensefull

Answer 2

Answer:

Step-by-step explanation:

To find the value of $x^{3}$ - 1/ $x^{3}$

x = 2 - $\sqrt{3}$

Step =

As x = 2 - $\sqrt{3}$

So 1/x = 1/2 - $\sqrt{3}$

Now rationalize the denominator of number 1/2 - $\sqrt{3}$ so that we could easily proceed

To rationalize the denominator of the number 1/2 - $\sqrt{3}$

Multiply both the denominator and numerator of the number with the conjugate of the denominator(2 - $\sqrt{3}$)

The conjugate o the denominator(2 - $\sqrt{3}$) is 2 + $\sqrt{3}$

That is,

1/x = 1/2 - $\sqrt{3}$ X 2 + $\sqrt{3}$/2 + $\sqrt{3}$

Follow the identity

1/x = 2 + $\sqrt{3}$/$(2)^{2}$ - $(\sqrt{3}) ^{2}$[Using identity $a^{2}$ - $b^{2}$ = (a + b)(a - b)

1/x = 2 + $\sqrt{3}$/4 - 3[ $(\sqrt{3}) ^{2}$ = 3]

1/x = 2 + $\sqrt{3}$(or 2 + $\sqrt{3}$/1)

Now we know

x = 2 - $\sqrt{3}$

1/x =2 + $\sqrt{3}$

To find the value of $x^{3}$ + 1/ $x^{3}$

$(x)^{3}$ - $(1/x)^{3}$= $(2 - \sqrt{3})^{3}$ - $(2 + \sqrt{3})^{3}$[Because x = 2 - $\sqrt{3}$ and 1/x =2 + $\sqrt{3}$]

Follow the identity

$(x)^{3}$ + $(1/x)^{3}$ = (2 - $\sqrt{3}$ - 2 + $\sqrt{3}$)($[2 - \sqrt{3}] ^{2}$ + (2 - $\sqrt{3}$)(2 + $\sqrt{3}$) + $[2 + \sqrt{3}] ^{2}$)[Using identity $a^{3}$ - $b^{3}$ = (a - b)($a^{2}$ + ab + $b^{2}$)]

$(x)^{3}$ + $(1/x)^{3}$ = (2 - 2)(4 - 4$\sqrt{3}$ + 3 + 4 + 2$\sqrt{3}$ - 2$\sqrt{3}$ - $\sqrt{9}$ + 4 + 4$\sqrt{3}$ + 3)

$(x)^{3}$ + $(1/x)^{3}$ = (0)(7 - 4$\sqrt{3}$ + 4 - 3( $\sqrt{9}$ = 3) + 7 + 4$\sqrt{3}$)

$(x)^{3}$ + $(1/x)^{3}$ = (0)(7 + 1 + 7)

$(x)^{3}$ + $(1/x)^{3}$ = 0 X 15

We know that anything multiplied by 0 is 0

So,

$(x)^{3}$ + $(1/x)^{3}$ = 0