If $x+\frac{1}{x}=a,x^2+\frac{1}{x^3}=b,$ then $x^3+\frac{1}{x^2}$ is _____

 $x+\frac{1}{x}=a\text{ and }{x}^2+\frac{1}{x^3}=b$
 ${(x+\frac{1}{x})}^2=a^2\Rightarrow x^2+\frac{1}{x^2}+2=a^2\mathrm{.........}\left(1\right)$
And ${(x+\frac{1}{x})}^3=a^3\Rightarrow x^3+\frac{1}{x^3}+3x.\frac{1}{x}(x+\frac{1}{x})=a^3\mathrm{............}\left(2\right)$  
Adding (1) & (2)
$(x^2+\frac{1}{x^3})+(x^3+\frac{1}{x^2})+2+3(x+\frac{1}{x})=a^2+a^3$ 
$b+(x^3+\frac{1}{x^2})+2+3a=a^2+a^3$ 
$\therefore x^3+\frac{1}{x^2}=a^2+a^3-3a-2-b$