Question

if x-1/x=1/3then find,x^3-1/x^3​

Answer 1

Answer:

${\sf{{\dfrac{19}{27}}}}$

Step-by-step explanation:

Given : ${\sf{\ \ x - {\dfrac{1}{x}} = {\dfrac{1}{3}} }}$

To Find : ${\sf{\ \ x^3 - {\dfrac{1}{x^3}}}}$

Solution :

Squaring both sides of the given equation.

${\sf{\Rightarrow \Big( x - {\dfrac{1}{x}} \Big) ^2 = \Big( {\dfrac{1}{3}} \Big) ^2}}$

${\boxed{\sf{\red{Identity \ : \ (a - b)^2 = a^2 + b^2 - 2ab}}}}$

${\sf{\red{Here, \ a = x, \ b = {\dfrac{1}{x}} }}}$

${\sf{\Rightarrow x^2 + {\dfrac{1}{x^2}} - 2.x.{\dfrac{1}{x}} = {\dfrac{1}{9}} }}$

${\sf{\Rightarrow x^2 + {\dfrac{1}{x^2}} - 2.{\cancel{x}}.{\dfrac{1}{{\cancel{x}}}} = {\dfrac{1}{9}} }}$

${\sf{\Rightarrow x^2 + {\dfrac{1}{x^2}} - 2 = {\dfrac{1}{9}} }}$

${\sf{\Rightarrow x^2 + {\dfrac{1}{x^2}} = {\dfrac{1}{9}} + 2}}$

${\sf{\Rightarrow x^2 + {\dfrac{1}{x^2}} = {\dfrac{1 + 18}{9}} }}$

${\sf{\Rightarrow x^2 + {\dfrac{1}{x^2}} = {\dfrac{19}{9}} }}$

Now,

${\boxed{\sf{\red{Identity \ : \ a^3 - b^3 = (a - b)(a^2 + b^2 + ab)}}}}$

${\sf{\red{Here, \ a = x, \ b = {\dfrac{1}{x}} }}}$

${\sf{\Rightarrow x^3 - {\dfrac{1}{x^3}} = \Big( x - {\dfrac{1}{x}} \Big) \Big( x^2 + {\dfrac{1}{x^2}} + x^2.{\dfrac{1}{x^2}} \Big) }}$

${\sf{\Rightarrow x^3 - {\dfrac{1}{x^3}} = \Big( x - {\dfrac{1}{x}} \Big) \Big( x^2 + {\dfrac{1}{x^2}} + {\cancel{x^2}}.{\dfrac{1}{{\cancel{x^2}}}} \Big) }}$

${\sf{\Rightarrow x^3 - {\dfrac{1}{x^3}} = \Big( x - {\dfrac{1}{x}} \Big) \Big( x^2 + {\dfrac{1}{x^2}} \Big) }}$

Putting known values.

${\sf{\Rightarrow x^3 - {\dfrac{1}{x^3}} = \Big( {\dfrac{1}{3}} \Big) \Big( {\dfrac{19}{9}} \Big) }}$

${\sf{\green{\Rightarrow x^3 - {\dfrac{1}{x^3}} = {\dfrac{19}{27}} }}}$