MHB Solve the cubic equation 4u^3+3u−3=0

[lfdahl]

Given the depressed cubic equation:

$4u^3+3u-3 =0$(1). Solve the equation with the substitution: $u = \sinh x$.

(2). Solve the equation without this substitution.

 

[I like Serena]

My solution:

Spoiler

(1)

\(
4u^3+3u-3=4\sinh^3 x+3\sinh x -3
=\frac 4{2^3}(e^{x}-e^{-x})^3+\frac 32(e^{x}-e^{-x}) -3 \\
=\frac 12(e^{3x}-3e^{x}+3e^{-x}-e^{-3x}) + \frac 32(e^{x}-e^{-x}) -3
=\frac 12(e^{3x}-e^{-3x}) -3 = \sinh(3x) - 3 =0
\)

So $x=\frac 13\arsinh(3)$.
And $u=\sinh(\frac 13\arsinh(3))\approx 0.644$.(2)

Define $u=\frac 12(x+y)$.
Then:

\(
4u^3+3u-3
=\frac{4}{2^3}(x+y)^3+\frac 32(x+y)-3
=\frac 12(x^3+y^3+3xy(x+y)) + \frac 32 (x+y) - 3 \\
=\frac 12(x^3+y^3-6) + \frac 32 (xy + 1)(x+y)
= 0
\)

We have the freedom to pick $y$ arbitrarily and we pick $y$ such that \( xy+1=0 \Rightarrow y=-\frac 1x \).
It follows that:

\( x^3+y^3-6=x^3-x^{-3}-6=0 \Rightarrow (x^3)^2-6(x^3)-1=0
\Rightarrow x^3=3\pm\sqrt{10} \\
\Rightarrow x=\sqrt[3]{\sqrt{10}+3}\lor x=-\sqrt[3]{\sqrt{10}-3}
\)

Since the same solutions will hold if we had swapped $x$ and $y$, we can conclude that we can pick $x$ to be the first solution and $y$ to be the second solution.
Substituting gives us \( u=\frac 12\left(\sqrt[3]{\sqrt{10}+3} - \sqrt[3]{\sqrt{10}-3}\right) \approx 0.644\).I see some similarity between both methods and I'm still wondering how close they are. (Thinking)

 

[lfdahl]

My solution:

Spoiler

(1)

\(
4u^3+3u-3=4\sinh^3 x+3\sinh x -3
=\frac 4{2^3}(e^{ix}-e^{-ix})^3+\frac 32(e^{ix}-e^{-ix}) -3 \\
=\frac 12(e^{3ix}-3e^{ix}+3e^{-ix}-e^{-3ix}) + \frac 32(e^{ix}-e^{-ix}) -3
=\frac 12(e^{3ix}-e^{-3ix}) -3 = \sinh(3x) - 3 =0
\)

So $x=\frac 13\arsinh(3)$.
And $u=\sinh(\frac 13\arsinh(3))\approx 0.644$.(2)

Define $u=\frac 12(x+y)$.
Then:

\(
4u^3+3u-3
=\frac{4}{2^3}(x+y)^3+\frac 32(x+y)-3
=\frac 12(x^3+y^3+3xy(x+y)) + \frac 32 (x+y) - 3 \\
=\frac 12(x^3+y^3-6) + \frac 32 (xy + 1)(x+y)
= 0
\)

We have the freedom to pick $y$ arbitrarily and we pick $y$ such that \( xy+1=0 \Rightarrow y=-\frac 1x \).
It follows that:

\( x^3+y^3-6=x^3-x^{-3}-6=0 \Rightarrow (x^3)^2-6(x^3)-1=0
\Rightarrow x^3=3\pm\sqrt{10} \\
\Rightarrow x=\sqrt[3]{\sqrt{10}+3}\lor x=-\sqrt[3]{\sqrt{10}-3}
\)

Since the same solutions will hold if we had swapped $x$ and $y$, we can conclude that we can pick $x$ to be the first solution and $y$ to be the second solution.
Substituting gives us \( u=\frac 12\left(\sqrt[3]{\sqrt{10}+3} - \sqrt[3]{\sqrt{10}-3}\right) \approx 0.644\).I see some similarity between both methods and I'm still wondering how close they are. (Thinking)

Brilliant, I like Serena! Your solution in (2). is new to me. Very nice indeed. Thankyou for your participation!