Question

if x minus one by x is equal to 1 by 3 evaluate x cube minus one by x cube

Answer 1

Answer:

$x - \frac{1}{x} = \frac{1}{3} \\ squaring \: on \: both \: sides \\ {(x - \frac{1}{x}) }^{2} = { \frac{1}{ {3}^{2} } }^{2} \\ {x}^{2} + \frac{1}{ {x}^{2} } - 2 = \frac{1}{9} \\ {x}^{2} + \frac{1}{ {x}^{2} } = \frac{1}{9} + 2 \\ {x}^{2} + \frac{1}{ {x}^{2} } = \frac{1 + 18}{9} = \frac{19}{9}$

according to question

${x}^{3} - \frac{1}{ {x}^{3} } = (x - \frac{1}{x} )( {x}^{2} + \frac{1}{ {x}^{2} } + 1) \\ = \frac{1}{3} ( \frac{19}{9} + 1) \\ = \frac{1}{3} ( \frac{19 + 9}{9} ) \\ = \frac{1}{3} ( \frac{28}{9} ) \\ = \frac{28}{27}$

Answer 2

The value of $x^{3}-\frac{1}{x^{3}}$ is $\frac{28}{27}$

Explanation:

It is given that $x-\frac{1}{x} =\frac{1}{3}$

To determine the value of $x^{3}-\frac{1}{x^{3}}$ , let us substitute the values in the formula,

$(a-b)^{3}=a^{3}-b^{3}-3 a b(a-b)$

Let us substitute $a=x$ and $b=\frac{1}{x}$

Thus, we get,

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

Substituting $x-\frac{1}{x} =\frac{1}{3}$ , in the above expression, we have,

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

Simplifying, we get,

$\frac{1}{27} =x^3-\frac{1}{x^3}-1$

Adding both sides of the equation by 1, we get,

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

     $\frac{28}{27}=x^{3}-\frac{1}{x^{3}}$

Thus, the value of $x^{3}-\frac{1}{x^{3}}$ is $\frac{28}{27}$

Learn more:

(1) if x minus one by x =3 evaluate x cube minus one by x cube

brainly.in/question/10388950

(2) If x minus one upon x is equal to 3 then find the value of x cube minus one by x cube equal to

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