Solve the cubic 1.5708x^3 - 0.01446x^2 + 1.35803e-5 x + 1.0159097e-12 = 0.

Step-by-step solution

1. 1

   Read the scale off the coefficients before touching a solver. The coefficients drop by roughly two decades per degree ($1.57$, $1.4\times10^{-2}$, $1.4\times10^{-5}$), which is the fingerprint of roots near $10^{-2}$ to $10^{-3}$. The constant term is a further seven decades smaller, so one root must be far tinier than the other two.

2. 2

   Rescale to a well-conditioned cubic. Substituting $x=t/1000$ multiplies the $t^k$ coefficient by $1000^{3-k}$:

   $1.5708t^3-14.46t^2+13.5803t+1.0159097\times10^{-3}=0$

   Now every coefficient sits within a couple of orders of magnitude, so a root finder loses far less precision to cancellation.

3. 3

   Locate the two large roots by sign changes. With $g(t)=1.5708t^3-14.46t^2+13.5803t+0.0010159097$: $g(1)=0.6921>0$, $g(1.5)=-6.8621<0$, $g(8)=-12.5470<0$, $g(9)=96.0769>0$. So roots lie in $(1,1.5)$ and $(8,9)$; bisecting to machine precision gives $t=1.06167$ and $t=8.14390$.

4. 4

   Get the tiny root from the constant term. Near $t=0$ the cubic and quadratic terms are negligible, so $13.5803t+0.0010159097\approx0$ gives $t\approx-7.4808\times10^{-5}$, refined by Newton to $t=-7.480165\times10^{-5}$. Dividing all three back by $1000$:

   $x=-7.480165\times10^{-8},\quad 0.00106167,\quad 0.00814390$

5. 5

   Check with Vieta's formulas, computed not assumed. Sum of roots should be $0.01446/1.5708=0.0092055004$; the three roots add to $0.0092055004$ ✓. Product should be $-1.0159097\times10^{-12}/1.5708=-6.467467\times10^{-13}$; the roots multiply to $-6.467467\times10^{-13}$ ✓. A widely circulated answer of $0.00100455$ and $0.00820134$ fails this test — those values leave residuals near $10^{-8}$ instead of $10^{-13}$.