# Cubic Equations

An cubic equation is an equation which can be represented in the form \(ax^3+bx^2+cx+d=0\), where \(a,b,c,d\) can be any number (may be complex if given so) where \(a \neq 0\).

An cubic equation has \(3\) roots (solutions), may be equal or not equal.

**Relation between coefficients and roots:**

For an cubic equation \(ax^3+bx^2+cx+d=0\), let \(p,q\) and \(r\) be its roots, then:

Root expression | Equals to |

\(p+q+r\) | \(-\frac{b}{a}\) |

\(pq+qr+rp\) | \(\frac{c}{a}\) |

\(pqr\) | \(-\frac{d}{a}\) |

Given that, \(p,q\) and \(r\) are its roots, then \((x-p)(x-q)(x-r)=0\) is the required cubic equation. Since it can be represented in the form \(ax^3+bx^2+cx+d=0 \implies x^3 + \frac{b}{a} x^2 + \frac{c}{a} x + \frac{d}{a} = 0\), we have the following approach:

\((x-p)(x-q)(x-r)=0\)

\((x-p)(x^2 - qx - rx + qr)=0\)

\((x^3 - qx^2 - rx^2 + qrx - px^2 + pqx + prx - pqr=0\)

\(x^3 + [-(p+q+r)]x^2 + (pq+qr+pr)x + (-pqr) = 0\)

Now comparing this with \(x^3 + (\frac{b}{a}) x^2 + (\frac{c}{a}) x + (\frac{d}{a}) = 0\), we have:

- \(p+q+r = -\frac{b}{a}\)
- \(pq+qr+pr = \frac{c}{a}\)
- \(pqr = -\frac{d}{a}\)

Find the sum of the squares of the roots of the cubic equation \(x^3 + 3x^2 + 3x = 3\).

Representing it in the form \(x^3 + 3x^2 + 3x - 3 = 0\).

Recall that \(p^2 + q^2 + r^2 = (p+q+r)^2 - 2(pq+qr+pr)\) (This is an algebraic identity and not related solely to cubic equations).

From the relations between the coefficients and its roots, we have \(p+q+r = -\frac{b}{a} = -3\) and \(pq+qr+pr = \frac{c}{a} = 3\).

Plugging it in the relation, we have \(p^2 + q^2 + r^2 = (p+q+r)^2 - 2(pq+qr+pr) = (-3)^2 - 2(3) = 9-6 = \boxed{3}\)

## Practice Problems