Question

if X square + 1 by x square equal to 7 then find the value of x cube + 1 by x q taking only the positive value of x + 1 by X

Answer 1

Answer:

Step-by-step explanation:

Answer 2

$\Large{\textbf{\underline{\underline{According\:to\:the\:Question}}}}$

$\textbf{\underline{Given}}$

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

${\boxed{\sf\:{Adding\;2\;to\;both\;sides\;we\;get :-}}}$

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

$\tt{\rightarrow (x+\dfrac{1}{x})^{2}=9}$

$\tt{\rightarrow x+\dfrac{1}{x}=\sqrt{9}}$

$\tt{\rightarrow x+\dfrac{1}{x}=\pm 3}$

$\textbf{\underline{Here}}$

$\textbf{\underline{Take\;plus\;value}}$

$\tt{\rightarrow x+\dfrac{1}{x}=3}$

$\textbf{\underline{Cubing\;both\;sides\;we\;get:-}}$

$\tt{\rightarrow (x+\dfrac{1}{x})^{3}=(3)^3}$

$\tt{\rightarrow x^3+\dfrac{1}{x^3}+3\times x\times\dfrac{1}{x}(x+\dfrac{1}{x})=27}$

$\tt{\rightarrow x^3+\dfrac{1}{x^3}+3\times 3=27}$

$\tt{\rightarrow x^3+\dfrac{1}{x^3}+9=27}$

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

$\tt{\rightarrow x^3+\dfrac{1}{x^3}=18}$

Hence,

${\boxed{\sf\:{x^3+\dfrac{1}{x^3}=18}}}$