How to solve the cubic equation $56z^3-70z^2-21z-4=0$?

Question

$56z^3-70z^2-21z-4=0$ how to solve for $z$. I formed three equations but not getting the answer. If I get a start or suggestion it would be a great help.

Answer 1

Here's a concise formula so anytime you come across a cubic you can easily solve it yourself. Given,

$$Ax^3+Bx^2+Cx+D=0$$

divide by $A$ to get the simpler,

$$x^3+ax^2+bx+c=0$$

then,

$$x = \frac{1}{3}\left(-a+u_1^{1/3}+\frac{a^2-3b}{u_1^{1/3}}\right)$$

where $u$ is any non-zero root of the quadratic,

$$u^2 + (2 a^3 - 9 a b + 27 c) u + (a^2 - 3 b)^3 = 0$$

P.S. This is the formula used by Mathematica though simplified for aesthetics.

Answer 2

HINT:

$$56z^3-70z^2-21z-4=0\Longleftrightarrow$$

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Eliminate the quadratic term by substituting $x=z-\frac{5}{12}$:

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$$56x^3-\frac{301x}{6}-\frac{563}{27}=0\Longleftrightarrow$$ $$x^3-\frac{43x}{48}-\frac{563}{1512}=0\Longleftrightarrow$$

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If $x=y+\frac{\lambda}{y}$ then $y=\frac{1}{2}\left(x+\sqrt{x^2-\lambda}\right)$:

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$$-\frac{563}{1512}-\frac{43}{48}\left(y+\frac{\lambda}{y}\right)+\left(y+\frac{\lambda}{y}\right)^3=0\Longleftrightarrow$$

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Substitute $\lambda=\frac{43}{144}$ and then $u=y^3$, yielding a quadratic equation in the variable $u$:

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$$u^2-\frac{563u}{1512}+\frac{79507}{2985984}=0$$