यदि $x^{2}+x=5$ तो $(x+3)^{3}+\frac{1}{(x+3)^{3}}$ का मूल्य है

$x^{2}+x=5$ (Given)

Let, $x+3=a$

$\therefore \quad \frac{1}{x+3}=\frac{1}{a}$

अभी,

$a+\frac{1}{a}=(x+3)+\frac{1}{(x+3)}$

$=\frac{(x+3)^{2}+1}{x+3}=\frac{x^{2}+6 x+9+1}{x+3}$

$=\frac{x^{2}+6 x+10}{x+3}=\frac{x^{2}+x+5 x+10}{x+3}$

$=\frac{5+5 x+10}{x+3}=\frac{5 x+15}{x+3}=\frac{5(x+3)}{x+3}=5$

$\therefore a^{3}+\frac{1}{a^{3}}$

$=\left(a+\frac{1}{a}\right)^{3}-3 a \times \frac{1}{a}\left(a+\frac{1}{a}\right)$

$=(5)^{3}-3 \times 5=125-15=110$