Solve x^3 + 12x^2 + 48x + 72 = 0.

Step-by-step solution

1. 1

   Test the perfect-cube pattern. The coefficients $1,12,48$ match $(x+b)^{3}=x^{3}+3bx^{2}+3b^{2}x+b^{3}$ with $3b=12$, so $b=4$; then $3b^{2}=48$ ✓. The only mismatch is the constant:

   $(x+4)^{3}=x^{3}+12x^{2}+48x+64,$

   whereas the problem has $72$. Recognising the near-miss is the whole idea — the cubic is not a perfect cube, but it is one plus a constant.

2. 2

   Rewrite as a shifted cube. Since $72=64+8$,

   $x^{3}+12x^{2}+48x+72=(x+4)^{3}+8,$

   so the equation is

   $(x+4)^{3}+8=0\;\Longrightarrow\;(x+4)^{3}=-8.$

   This single rewrite converts a cubic into something solvable in one step.

3. 3

   Take the real cube root. Unlike a square root, a cube root of a negative number is a perfectly good real number, and it is unique:

   $x+4=\sqrt[3]{-8}=-2.$

   There is no $\pm$ here — the function $t\mapsto t^{3}$ is strictly increasing, so it is one-to-one on the reals.

4. 4

   Solve for $x$. Subtracting $4$:

   $x=-2-4=-6.$

   This is the unique real root; the other two are complex, coming from the factorisation $(x+4)^{3}+2^{3}=(x+6)\left((x+4)^{2}-2(x+4)+4\right)$ via the sum-of-cubes identity, whose quadratic factor $x^{2}+6x+12$ has discriminant $36-48=-12<0$.

5. 5

   Check by substitution. With $x=-6$:

   $(-6)^{3}+12(36)+48(-6)+72=-216+432-288+72=0 ✓.$

   A structural check also works: $(-6+4)^{3}+8=(-2)^{3}+8=-8+8=0$ ✓.