<s>[INST] How to write this cubic equation For an imaginary number $i=\sqrt{-1}$ ,the cubic equation $24x^3+21x^2-72x-7=0$ can be represented in the form $$\frac{((x+i)^3-(x-i)^3)}{((x+i)^3+(x-i)^3)}\frac{((2+i)^4+(2-i)^4)}{((2+i)^4-(2-i)^4)}=1$$

How can we write the cubic $141x^3+72x^2-141x-8=0$  in a similar form? 

 [/INST] HINT:
By applying Componendo and Dividendo repeatedly,
we find $$\dfrac{x^3-3x}{-7}=\dfrac{3x^2-1}{24}\iff24x^3+21x^2-72x-7=0$$
Now,
$$14x1^3+72x^2-141x-8=0\iff\dfrac{x^3-x}8=\dfrac{1-9x^2}{141}$$
Again, $(ax+i)^3=a^2x^3+i(3a^2x^2)+3ax(-1)-i$
Comparing with  $x^3-x+i(1-9x^2)$
we need, $$\dfrac{a^3}1=\dfrac{3a}1\iff a^3=3a$$
As $a\ne0,a=\pm\sqrt3$
Taking $a=\sqrt3,$
$$\dfrac{3\sqrt3x^3-\sqrt3x}{24\sqrt3}=\dfrac{(1-9x^2)}{141}$$
$$\iff\dfrac{3\sqrt3x^3-\sqrt3x}{(1-9x^2)i}=\dfrac{-24\sqrt3i}{141}$$
Applying Componendo and Dividendo,  $$\left(\dfrac{i+\sqrt3x}{i-\sqrt3x}\right)^3=\dfrac{141+24\sqrt3i}{141-24\sqrt3i}$$
I leave to you to express $$\dfrac{141+24\sqrt3i}{141-24\sqrt3i}$$
as $$\left(\dfrac{c+id}{c-id}\right)^4$$
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