There are three distinct real numbers $a, b$ and $c$ that are solutions of the equation $x^{3}-4 x=0$. What is the value of the product $a b c$ ?
Solution 1

Since $x^{3}-4 x=0$, then $x\left(x^{2}-4\right)=0$ which is equivalent to $x(x+2)(x-2)=0$.

Therefore, the solutions to the equation are $x=0,-2,2$.

These three numbers are $a, b$ and $c$ in some order.

Therefore, $a b c=0(-2)(2)=0$.

(Note that once we determined that one of the solutions was 0 , we did not need to do any more work, since the product of the roots is 0 .)

## Solution 2

When a cubic equation has the form $x^{3}-S x^{2}+T x-P=0$, then the product of the solutions equals $P$.

In this problem, the cubic equation has the form $x^{3}-0 x^{2}+(-4) x-0=0$.

Since the solutions are $a, b$ and $c$, then $P=a b c$.

But $P=0$, so $a b c=0$.

ANSWER: 0

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