Write 27x^3 - 54x^2 + 36x - 8 = 0 using a binomial identity and solve it.

Step-by-step solution

1. 1

   Check the two end terms for perfect cubes. A cubic can only be $(a-b)^3$ if its first and last terms are cubes:

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

   so the candidate is $a=3x$, $b=2$, and the alternating signs point to the minus version of the identity.

2. 2

   Test the two middle terms against the identity. The pattern is

   $(a-b)^3=a^3-3a^2b+3ab^2-b^3$

   With $a=3x$ and $b=2$:

   $3a^2b=3\left(9x^2\right)(2)=54x^2\;\checkmark,\qquad 3ab^2=3(3x)(4)=36x\;\checkmark$

   Both match, so the identification is confirmed rather than merely plausible.

3. 3

   Write the equation in cube form.

   $27x^3-54x^2+36x-8=(3x-2)^3=0$

4. 4

   Solve. A cube is zero only when its base is zero:

   $3x-2=0\;\Longrightarrow\;x=\frac{2}{3}$

   This is a triple root: one distinct solution with multiplicity $3$. Consequently the graph flattens against the axis at $x=\tfrac23$ instead of cutting through steeply.

5. 5

   Verify by substitution. At $x=\tfrac23$: $27\cdot\tfrac{8}{27}-54\cdot\tfrac49+36\cdot\tfrac23-8=8-24+24-8=0\;\checkmark$. A second spot check at $x=1$ gives $27-54+36-8=1$ against $(3-2)^3=1\;\checkmark$.