Single Variable Calculus, Chapter 1, 1.2, Section 1.2, Problem 9

Suppose that for a cubic function of $f, f(1) = 6$ and $f(-1) = f(0) =f(2)=0$. Find its expression.


$
\begin{equation}
\begin{aligned}
y = ax^3+bx^2+cx+d\\
@ \, f(1) &= 6\\
&a(1)^3 + b(1)^2 + c(1) + d &=6\\
&a+b+c+d &= 6 &&\text{Equation 1}\\

@ f(-1) &=0\\
&a(-1)^3+b(-1)^2+c(-1)+d &= 0\\
&-a+b-c+d &= 0 &&\text{Equation 2}\\
@f(0) &= 0\\
&a(0)^3+b(0)^2+c(0)+d&=0\\
&d&=0 &&\text{Equation 3}\\
@f(2) &= 0\\
&a(2)^3+b(2)^2+c(2)+d &= 0\\
&8a+4b+2c+d&=0 &&\text{Equation 4}
\end{aligned}
\end{equation}
$



*Combining equations 1 to 4 will result to

$a = -3, \, b = 3, \, c=6$

$\boxed{.: \text{the cubic function is } -3x^3+3x^2+6x = 0}$