An identity relating the roots of $F(x)=x^3+x^2+4x+4$

Question

Demonstrate how $$\frac1{x_1} + \frac1{x_2} + \frac1{x_3} + \frac1{x_1x_2} + \frac1{x_2x_3} + \frac1{x_3x_1} = -\frac34$$ where $x_1, x_2, x_3$ are roots of the polynomial $F(x) = x^3 + x^2 + 4x + 4$.

Can someone help me please, thank you!

Answer 1

Hint:

Use Vieta's formulas. The result should be easy after.

Answer 2

Here's a hint which doesn't use Vieta explicitly: Since $x_1,x_2,x_3$ are roots of $F(x)=x^3+x^2+4x+4$, you can write $F(x)=(x-x_1)(x-x_2)(x-x_3)$. Try evaluating $F(x)$ at $x=0$ and $x=-1$, in each case using both this form of $F(x)$ and the original form.

Answer 3

Not sure if I am missing something obvious here, but can't the given polynomial be factored by grouping? Just find the roots explicitly and calculate the equation.

Answer 4

The LHS is $$\frac{(x_2x_3+x_1x_3+x_1x_2)+(x_3+x_1+x_2)}{x_1x_2x_3}=\frac{(4)+(-1)}{-4}$$