Answer
$0.819172513$
Work Step by Step
Step 1. Letting $f(x)=x^2(x+1)-\frac{1}{x}=x^3+x^2-\frac{1}{x}, x\gt0$, we have $f'(x)=3x^2+2x+\frac{1}{x^2}$. Finding the intersect of the graphs given in the exercise is the same as finding the zero of $f(x)$.
Step 2. Test signs of $f(x)$, $f(0.1)=0.01(1.1)-10\lt0$ and $f(1)=1(2)-1=1\gt0$; thus there is a zero in the interval of $(0.1,1)$
Step 3. Using Newton’s method $x_{n+1}=x_n-\frac{f(x_n)}{f'(x_n)}$ with a starting point $x_0=1$, we can obtain the zero shown in the table as $0.819172513$
