Question 1184166
.
Solve algebraic {{{cross(expression)}}} &nbsp;&nbsp;<U>EQUATION</U> &nbsp;&nbsp;{{{ a^6 -12a^4 + 48a^2 - 61 = 0 }}} &nbsp;&nbsp;and find the real roots.
~~~~~~~~~~~~~~~~~~~



&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;It looks as a miracle, &nbsp;but I will give a precise solution to the given equation.



<pre>
The starting equation is

    {{{ a^6 -12a^4 + 48a^2 - 61}}} = {{{0 }}}.       (1)



Recall the standard formula  (x-y)^3 = x^3 - 3x^2y + 3x*y^2 - y^3.



In the formula, put  x = a^2,  y = 4.  You will get

    {{{a^6 -12a^4 + 48a^2 - 61}}} = {{{a^6 -12a^4 + 48a^2 - 64}}} + 3.



Hence, the given equation is EQUIVALENT to

    {{{(a^2-4)^3}}} = -3.      (2)


It implies that     


    {{{a^2-4}}} = {{{root(3,-3)}}},

or

    {{{a^2-4}}} = - {{{root(3,3)}}}.    (3)



Hence,  

    {{{a^2}}} = {{{4 - root(3,3)}}},

    a = +/- {{{sqrt(4-root(3,3))}}}.     (4)


Approximate numerical values of the roots (4) are  +/- 1.599297.
</pre>

Solved.