Find zeros of a polynomial and apply them to an another equation

Question

If $a$ and $b$ are the zeros of quadratic polynomial $x^2-4x-5$, find the value of $\frac1{a^3} + \frac1{b^3}$.

Answer 1

Another way to approach it without solving the corresponding equation. One has \begin{align*} \frac{1}{a^{3}} + \frac{1}{b^{3}} = \frac{a^{3}+b^{3}}{a^{3}b^{3}} = \frac{(a+b)(a^{2} - ab + b^{2})}{a^{3}b^{3}} = \frac{(a+b)[(a+b)^{2} - 3ab]}{a^{3}b^{3}} \end{align*}

Since $a + b = 4$ and $ab = -5$, we conclude that \begin{align*} \frac{1}{a^{3}} + \frac{1}{b^{3}} = -\frac{4\times(4^{2} + 15)}{125} = -\frac{124}{125} \end{align*}

Hopefully this helps.

Answer 2

We know that: $$x^2-4x-5=(x-5)(x+1)$$ The roots are: $$x=5 \vee x=-1$$ Now, we can compute the value of: $$\frac{1}{a^3}+\frac{1}{b^3}=\frac{1}{5^3}-1=-\frac{124}{125}$$