Solve 27x^3 - 54x^2 + 36x - 8 = 0.

Step-by-step solution

1. 1

   Look for a perfect cube before reaching for the cubic formula. The first and last terms are both perfect cubes:

   $27x^{3}=(3x)^{3},\qquad 8=2^{3}.$

   Alternating signs ($+,-,+,-$) point to the difference form $(a-b)^{3}$, so it is worth testing $a=3x$ and $b=2$ before doing anything harder. Recognising the pattern turns a cubic into a one-line problem.

2. 2

   Expand $(3x-2)^{3}$ and compare. Using $(a-b)^{3}=a^{3}-3a^{2}b+3ab^{2}-b^{3}$:

   $(3x-2)^{3}=(3x)^{3}-3(3x)^{2}(2)+3(3x)(2)^{2}-2^{3}=27x^{3}-54x^{2}+36x-8.$

   Every coefficient matches the original. Check the middle two carefully: $3\cdot 9x^{2}\cdot 2=54x^{2}$ and $3\cdot 3x\cdot 4=36x$ — these are the terms that would fail if the guess were wrong.

3. 3

   Rewrite the equation in factored form. The equation is therefore

   $(3x-2)^{3}=0.$

   A cube equals zero only when its base does, since $t^{3}=0\iff t=0$ over the reals and the complex numbers. There is no second case to consider.

4. 4

   Solve the linear equation. From $3x-2=0$:

   $3x=2\;\Longrightarrow\;x=\frac{2}{3}.$

   So the cubic has the single root $x=\tfrac23$, with multiplicity three — all three roots of this degree-3 polynomial coincide. Graphically the curve flattens against the $x$-axis at that point rather than crossing it steeply.

5. 5

   Check by substitution and by Vieta. Substituting: $27\cdot\tfrac{8}{27}-54\cdot\tfrac49+36\cdot\tfrac23-8=8-24+24-8=0$ ✓. Vieta also agrees: the three roots must sum to $-\tfrac{-54}{27}=2=3\times\tfrac23$ ✓ and multiply to $\tfrac{8}{27}=\left(\tfrac23\right)^{3}$ ✓.