Question

if x2+1/x2=27 then find the value of x3_ 1/x3​

Answer 1

Step-by-step explanation:

Given:-

$\sf x^2+ \dfrac{1}{x^2}=27$

To find:-

$\sf x^3- \dfrac{1}{x^3}$

Solution:-

$\sf First \: we \: will \: find \: x - \dfrac{1}{x}$ $\sf \: to \: find \: x^3 - \dfrac{1}{x^3}$

$\sf We \: know \: that, \\ \\ \sf (a-b)^2 = a^2+b^2-2ab$

$\sf Using \: this \: identity:- \\ \\ \sf \implies (x- \dfrac{1}{x})^2= x^2+ \dfrac{1}{x^2} - 2 \times x \times \dfrac{1}{x}$

$\sf \implies (x- \dfrac{1}{x})^2=27-2 \\ \\ \sf \implies (x- \dfrac{1}{x})^2=25 \\ \\ \sf \implies x- \dfrac{1}{x} = \sqrt{25} \\ \\ \sf \implies x- \dfrac{1}{x}= 5$

$\sf Now, \: we \: will \: find \: x^3 - \dfrac{1}{x^3} \\ \\ \sf We \: will \: use \: the \: identity \implies (a-b)^3 = a^3 - b^3 - 3ab(a-b)$

$\sf \: to \: find \: x^3 - \dfrac{1}{x^3}$

$\sf \implies (x - \dfrac{1}{x})^3 = (x)^3 - (\dfrac{1}{x})^3 - 3 \times x \times \dfrac{1}{x} (x- \dfrac{1}{x})$

$\sf \implies (5)^3= x^3 - \dfrac{1}{x^3} - 3(5) \\ \\ \sf \implies 125= x^3 - \dfrac{1}{x^3} - 15 \\ \\ \sf \implies 125+15 = x^3 - \dfrac{1}{x^3} \\ \\ \sf \implies 140 = x^3 - \dfrac{1}{x^3}$