Question

please solve sum 21, I mark him as brianliest

Answer 1

First, square x+$\frac{1}{x}$
doing this, we get,
 $x^{2} +[tex]$ \frac{1}{x^{2}} +2[/tex]
which is just 2 more than the given value.
Hence, $(x+ \frac{1}{x}) ^{2} =25$
or, $(x+ \frac{1}{x})=5$

Next,
 (x+ \frac{1}{x} )^{3}= x^{3}+ \frac{1}{x^{3}} +3(x+ \frac{1}{x})
or, $x^{3}+ \frac{1}{x^{3}} =5^{3}-3(5)$
therefore, $x^{3}+ \frac{1}{x^{3}}=125-15=110$