Question

if x(3-2/x)=3/x then x3-1/x3=?

Answer 1

3x-1/3x=0
3x-1=0
3x=1
x=1/3

Answer 2

Answer: Our required value would be

$x^3-\dfrac{1}{x^3}=\dfrac{62}{27}$

Step-by-step explanation:

Since we have given that

$x(3-\dfrac{2}{x})=\dfrac{3}{x}\\\\3x-2=\dfrac{3}{x}\\\\3x-\dfrac{3}{x}=2\\\\3(x-\dfrac{1}{x})=2\\\\x-\dfrac{1}{x}=\dfrac{2}{3}$

We need to find the value of $x^3-\dfrac{1}{x^3}$

As we know that

$x^3-\dfrac{1}{x^3}=A^3+3A=(\dfrac{2}{3})^3+3\times \dfrac{2}{3}=\dfrac{8}{27}+2=\dfrac{8+54}{27}=\dfrac{62}{27}$

So, our required value would be

$x^3-\dfrac{1}{x^3}=\dfrac{62}{27}$