Question

If x + 1/x = 10/3 find x3 - 1/3x

Answer 1

Answer:

$x + \frac{1}{x} =\frac{10}{3} \\\frac{x^{2} + 1}{x} = 10/3\\3(x^{2} + 1) = 10x\\3x^{2} -10x +3 = 0\\\\x = 9 (quadratic formula)\\x^{3} -\frac{1}{3x} = 9^{3} - \frac{1}{3(9)} \\= 729 - \frac{1}{27} = \frac{19683-1}{27} = \frac{19682}{27}$

Step-by-step explanation:

Answer 2

__________________________________________________

Question :

$\sf \small If \: x + \frac{1}{ x } = \frac{10}{3} \: then \: find \: {x}^{3} - \frac{1}{ {x}^{3} }$

To find :

${x}^{3} - \frac{1}{ {x}^{3} }$

Solution :

First we need to get x² + 1/x² to get x - 1/x and then we need to cube x - 1/x to get the value of x³ - 1/x³

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

${( \frac{10}{3}) }^{2} = {x}^{2} + \frac{1}{ {x}^{2} } + 2$

${x}^{2} + \frac{1}{ {x}^{2} } = \frac{100}{9} - 2$

${x}^{2} + \frac{1}{ {x}^{2} } = \frac{100 - 18}{9}$

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

Now we will apply it in finding x - 1/x

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

By putting the value of x² + 1/x² we get

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

${(x - \frac{1}{x} )}^{2} = \frac{82 - 18}{9}$

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

$x - \frac{1}{x} = \sqrt{ \frac{64}{9} }$

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

According to the plan we now need to cube x - 1/x

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

Now we need to put the values in place of these variables

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

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

${x}^{3} - \frac{1}{ {x}^{3} } = \frac{8 + 24}{3}$

${x}^{3} - \frac{1}{ {x}^{3} } = \frac{32}{3}$

${x}^{3} - \frac{1}{ {x}^{3} } \rarr 10 \frac{2}{3}$

$\sf \small \blue{So }\: \green{ the} \: \pink{value }\: \orange{is } \rarr \boxed{ \purple{ 10\frac{2}{3} }}$

__________________________________________________