Solve x^3 - 0.016875x^2 + 0.000094921875x - 1.77975547e-7 = 0.

Step-by-step solution

1. 1

   Test whether the first three terms are a perfect cube. For $(x-a)^3=x^3-3ax^2+3a^2x-a^3$ the $x^2$ coefficient gives $3a=0.016875$, so $a=0.005625$. Check the $x$ coefficient: $3a^2=3(0.005625)^2=3(0.000031640625)=0.000094921875$ — an exact match. This is not a coincidence; the equation was built from a shifted cube.

2. 2

   Rewrite the equation as a pure cube plus a constant. Since $a^3=(0.005625)^3=1.779785156\overline{25}\times10^{-7}$,

   $x^3-0.016875x^2+0.000094921875x=(x-0.005625)^3+a^3$

   so the equation becomes

   $(x-0.005625)^3=1.77975547\times10^{-7}-1.77978515625\times10^{-7}=-2.968625\times10^{-12}$

3. 3

   Take the real cube root. A real number has exactly one real cube root, so

   $x-0.005625=-\sqrt[3]{2.968625\times10^{-12}}=-1.4372041\times10^{-4}$

   $x=0.005625-0.00014372=0.00548128$

   The negative sign survives because the right-hand side is negative.

4. 4

   Account for the other two roots. The remaining roots come from the complex cube roots of unity, $x=a+r\,e^{\pm 2\pi i/3}$ with $r=-1.4372041\times10^{-4}$:

   $x\approx 0.00569686\pm 0.00012447\,i$

   So the cubic has one real root and a conjugate pair — it does not have a repeated real root, a claim that fails the discriminant test since $(x-a)^3=c$ has a triple root only when $c=0$.

5. 5

   Verify the real root by direct substitution. With $x=0.005481279590$: $x^3=1.6467\times10^{-7}$, $-0.016875x^2=-5.0698\times10^{-7}$, $+0.000094921875x=5.2030\times10^{-7}$, $-1.77975547\times10^{-7}$. Summing gives $-5.3\times10^{-23}$, i.e. zero to machine precision ✓. Substituting the frequently quoted $0.0056808311$ instead leaves a residual near $10^{-12}$, so that value is not a root.