If \(\alpha,\beta\) are the roots of the quadratic equation
\[
x^2-3x+1=0,
\]
then the value of \(\alpha^3+\beta^3\) is:
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For roots \(\alpha,\beta\) of a quadratic equation, always remember:
\[
\alpha+\beta=-\frac{b}{a},
\qquad
\alpha\beta=\frac{c}{a}.
\]
Many questions involving powers of roots can be solved without finding the actual roots.
Step 1: Read off the root relations. For $x^2-3x+1=0$, the sum of roots is $\alpha+\beta=3$ and the product is $\alpha\beta=1$. These come from comparing with $x^2-(\text{sum})x+(\text{product})=0$.
Step 2: Pick the right identity. We want $\alpha^3+\beta^3$. A clean identity is $\alpha^3+\beta^3=(\alpha+\beta)^3-3\alpha\beta(\alpha+\beta)$. This avoids finding the actual roots.
Step 3: Plug in the sum cubed. Compute $(\alpha+\beta)^3=3^3=27$.
Step 4: Plug in the product term. Compute $3\alpha\beta(\alpha+\beta)=3\times 1\times 3=9$.
Step 5: Subtract. So $\alpha^3+\beta^3=27-9=18$.
Step 6: Double-check the logic. The identity is just the expansion of $(\alpha+\beta)^3$ rearranged, so it is exact. No approximation is involved.