Question

if X square + 1 upon x is equal to 5 by 2 find the value of x minus one by x and x cube minus one by x cube ​

Answer 1

$\textbf{Given:}$

$\mathsf{\dfrac{x^2+1}{x}=\dfrac{5}{2}}$

$\textbf{To find:}$

$\textsf{The value of}\;\mathsf{(i)\;x-\dfrac{1}{x}\;\;(ii)\;x^3-\dfrac{1}{x^3}}$

$\textbf{Solution:}$

$\textbf{Formula used:}$

$\boxed{\begin{minipage}{7cm} \\\mathsf{1.\;(a-b)^2=(a+b)^2-4ab}\\\\\mathsf{2.\;(a-b)^3=a^3-b^3-3ab(a-b)} \end{minipage}}$

$\mathsf{Consider,}$

$\mathsf{\dfrac{x^2+1}{x}=\dfrac{5}{2}}$

$\mathsf{x+\dfrac{1}{x}=\dfrac{5}{2}}$

$\mathsf{Using\;formula\;(1)}$

$\mathsf{\left(x-\dfrac{1}{x}\right)^2=\left(x+\dfrac{1}{x}\right)^2-4}$

$\mathsf{\left(x-\dfrac{1}{x}\right)^2=\left(\dfrac{5}{2}\right)^2-4}$

$\mathsf{\left(x-\dfrac{1}{x}\right)^2=\dfrac{25}{4}-4}$

$\mathsf{\left(x-\dfrac{1}{x}\right)^2=\dfrac{25-16}{4}}$

$\mathsf{\left(x-\dfrac{1}{x}\right)^2=\dfrac{9}{4}}$

$\mathsf{x-\dfrac{1}{x}\right=\sqrt{\dfrac{9}{4}}}$

$\implies\boxed{\mathsf{x-\dfrac{1}{x}\right=\dfrac{3}{2}}}$

$\mathsf{Using\;formula\;(2),\;we\;get}$

$\mathsf{\left(x-\dfrac{1}{x}\right)^3=x^3-\dfrac{1}{x^3}-3\left(x-\dfrac{1}{x}\right)}$

$\mathsf{\left(\dfrac{3}{2}\right)^3=x^3-\dfrac{1}{x^3}-3\left(\dfrac{3}{2}\right)}$

$\mathsf{\dfrac{27}{8}=x^3-\dfrac{1}{x^3}-\dfrac{9}{2}}$

$\mathsf{\dfrac{27}{8}+\dfrac{9}{2}=x^3-\dfrac{1}{x^3}}$

$\mathsf{\dfrac{27+36}{8}=x^3-\dfrac{1}{x^3}}$

$\implies\boxed{\mathsf{x^3-\dfrac{1}{x^3}=\dfrac{63}{8}}}$

$\textbf{Find more:}$

if X square + 1 by x square is equal to 7 then find the value of x cube + 1 by x cube and x minus one upon x

https://brainly.in/question/7239360

If 5x – 4y =8 and xy=12, then find the value of (5x+4y)^2

https://brainly.in/question/15201072

Answer 2

Step-by-step explanation:

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