Question

if x - 1/x = 2 then what is x^3 - 1/x^3

Answer 1

Given,x-1/x=2
Using a^3-b^3=(a-b)^3+3(a)(b)(a-b)

x^3-1/x^3=(x-1/x)^3+3(x)(1/x)(x-1/x)
=(2)^3+3(1)(2)
=8+6
=14

Answer 2

$\bold{\large{\boxed{\red{\tt{ANSWER}}}}}$

It is given that ,

$\boxed{\bold{\blue{\tt{x - \frac{1}{x} = 2}}}} - - > eq \: 1$

Now cubing on both sides we get ,

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

$\rightarrow{ ({x - \frac{1}{x}) }^{3} = 8} - - > eq \: 2$

Now by using identity ,

$\bold{\boxed{\large{\pink{\tt{ {(a - b)}^{3} = {a}^{3} - {b}^{3} - 3ab(a - b)}}}}}$

$\rightarrow{{x}^{3} - \frac{1}{ {x}^{3} } - 3(x \times \frac{1}{x} )(x - \frac{1}{x} ) = 8 - - > (from \: eq \: 2)}$

$\rightarrow{{x}^{3} - \frac{1}{ {x}^{3} } - 3(x - \frac{1}{x} ) = 8}$

$\rightarrow{ {x}^{3} - \frac{1}{ {x}^{3} } - 3(2) = 8 - - > from \: eq \: 1}$

$\rightarrow{ {x}^{3} - \frac{1}{ {x}^{3} } - 6 = 8}$

$\rightarrow{x}^{3} - \frac{1}{ {x}^{3} } = 8 + 6$

$\rightarrow{\boxed{\bold{\green{\tt{ {x}^{3} - \frac{1}{ {x}^{3} } = 14}}}}}$

$\bold{\large{\purple{\tt{I \: hope \: it \: helps \: u \: dear \: friend !!}}}}$