If x-dfrac -1-x-sqrt 3- then x-3-dfrac -1-x-3-6sqrt 32sqrt 33sqrt 3-6sqrt 3

Question

Toppr Admin · Accepted Answer

X-x2212-1x-x2212-x221A-3Take cube of both sides- we get-x-x2212-1x-3-x2212-x221A-3-3-x21D2-x3-x2212-1x3-x2212-3-x22C5-x-x22C5-1x-x-x2212-1x-x2212-3-x221A-3-xA0-by using-xA0-a-x2212-b-3-a3-x2212-b3-x2212-3ab-a-x2212-b-x21D2-x3-x2212-1x3-x2212-3-x2212-x221A-3-x2212-3-x221A-3-x21D2-x3-x2212-1x3-3-x221A-3-x2212-3-x221A-3-x21D2-x3-x2212-1x3-x2212-6-x221A-3So- option D is correct-

If x−1x=−√3, then find x3−1x3.6√32√33√3−6√3

x−1x=−√3Take cube of both sides, we get(x−1x)3=(−√3)3⇒x3−1x3−3⋅x⋅1x(x−1x)=−3√3 by using (a−b)3=a3−b3−3ab(a−b)⇒x3−1x3−3(−√3)=−3√3⇒x3−1x3+3√3=−3√3⇒x3−1x3=−6√3So, option D is correct.

If $x - \frac{1}{x} = - \sqrt{3}$ , then find $x^{3} - \frac{1}{x^{3}}$ .
$6 \sqrt{3}$
$2 \sqrt{3}$
$3 \sqrt{3}$
$- 6 \sqrt{3}$