Solving related equations by substitution

Once a cubic \(f(x)=0\) has been solved, an equation that is the same cubic in a different expression can be solved without starting again. Replace that expression by a single letter, use the roots you already have, then solve for the new variable.

Why: if \(f(u)=0\) only when \(u\) is one of the roots \(r_1\), \(r_2\), \(r_3\), then \(f(\text{expression})=0\) only when the expression equals one of those roots.

Method

1. Compare the new equation with \(f(x)=0\) and find the expression \(u\) that has taken the place of \(x\) (for example \(u=y-1\), \(u=y^2\) or \(u=\dfrac1y\)). Rearrange or divide first if needed.
2. Write \(u=r_1\), \(u=r_2\), \(u=r_3\).
3. Solve each for the new variable.
4. Reject values that are impossible, such as \(y^2=-2\), and say why.

Example

\(x^3-2x^2-5x+6=(x-1)(x-3)(x+2)\), so \(x^3-2x^2-5x+6=0\) has roots \(1\), \(3\) and \(-2\).

Solve \((y-1)^3-2(y-1)^2-5(y-1)+6=0\). Here \(u=y-1\): \(y-1=1,\ 3\) or \(-2\), so \(y=2,\ 4\) or \(-1\).

Solve \(y^6-2y^4-5y^2+6=0\). Here \(u=y^2\): \(y^2=1\) or \(y^2=3\), so \(y=\pm1\) or \(y=\pm\sqrt3\); \(y^2=-2\) has no real solution.