Question

यदि x2+1/x2=82/9 तो x3-1/x3 का मान ज्ञात करो

Answer 1

Hope the solution attached will help you.

Answer 2

Answer:

$x^3+\frac{1}{x^3}=\frac{728}{27}$

Step-by-step explanation:

Given : $x^2+\frac{1}{x^2}=\frac{82}{9}$

To find : $x^3+\frac{1}{x^3}$ ?

Solution :

We know that,

$(x-\frac{1}{x})^2=x^2+\frac{1}{x^2}-2\times x\times \frac{1}{x}$

Substitute the given value,

$(x-\frac{1}{x})^2=\frac{82}{9}-2$

$(x-\frac{1}{x})^2=\frac{64}{9}$

Taking root both side,

$x-\frac{1}{x}=\frac{8}{3}$

Applying cubic formula,

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

$x^3+\frac{1}{x^3}=(\frac{8}{3})(\frac{82}{9}+1)$

$x^3+\frac{1}{x^3}=(\frac{8}{3})(\frac{91}{9})$

$x^3+\frac{1}{x^3}=\frac{728}{27}$

Therefore, $x^3+\frac{1}{x^3}=\frac{728}{27}$