Question

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

Answer 1

$\underline{\textsf{Given:}}$

$\mathsf{x^2+\dfrac{1}{x^2}=7}$

$\underline{\textsf{To find:}}$

$\textsf{The value of}$

$\mathsf{x^3+\dfrac{1}{x^3}}\;\textsf{and}$

$\mathsf{x-\dfrac{1}{x}}$

$\underline{\textsf{Solution:}}$

$\textsf{Consider,}$

$\mathsf{x^2+\dfrac{1}{x^2}=7}$

$\textsf{This can be writtten as}$

$\mathsf{x^2+\dfrac{1}{x^2}+2\,x(\dfrac{1}{x})=7+2}$

$\mathsf{(x+\frac{1}{x})^2=9}$

$\implies\mathsf{x+\frac{1}{x}=3}$

$\textsf{Using the identity,}$

$\mathsf{a^3+b^3=(a+b)^3-3ab(a+b)}$

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

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

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

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

$\implies\boxed{\mathsf{x^3+\dfrac{1}{x^3}=18}}$

$\textsf{We know that}$

$\mathsf{(a-b)^2=(a+b)^2-4ab}$

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

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

$\implies\mathsf{(x-\dfrac{1}{x})^2=(3)^2-4}$

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

$\implies\mathsf{(x-\dfrac{1}{x})^2=5}$

$\implies\boxed{\mathsf{x-\dfrac{1}{x}=\sqrt{5}}}$