We are given equation :[tex]x^6+6x^3+5=0[/tex]
[tex]Use\:the\:rational\:root\:theorem[/tex]
[tex]\mathrm{Therefore,\:we\:need\:to\:check\:the\:following\:rational\:numbers:\quad }\pm \frac{1,\:5}{1}[/tex]
[tex]-\frac{1}{1}\mathrm{\:is\:a\:root\:of\:the\:expression,\:so\:factor\:out\:}x+1[/tex]
[tex]\mathrm{Compute\:}\frac{x^6+6x^3+5}{x+1}\mathrm{\:to\:get\:the\:rest\:of\:the\:eqution:\quad }x^5-x^4+x^3+5x^2-5x+5[/tex]
Therefore, final factored form it
[tex]x^6\:+\:6x^3\:+\:5=\left(x+1\right)\left(x^5-x^4+x^3+5x^2-5x+5\right)[/tex]
We can't factor it more.
Therefore,
x+1=0.
x=-1.
Therefore, the real solution of the equation would be -1.
The equation [tex]x^6+6x^3+5=0[/tex] which is a sixth order equation can be solved easily by making a substitution that turns the equation into an equivalent quadratic equation.
We will make the substitution, [tex]x^3=y[/tex] , to simplify the equation. The two roots of this equation are [tex]x=-1,x=-5^{{1}/{3}}[/tex]
[tex]x^6+6x^3+5=(x^3)^2+6x^3+5=y^2+6y+5=0[/tex].
The next step is to solve for [tex]y[/tex] in the equation on the right by factorization. We will look for two factors of 5 that add up to 6 and then use those factors to factorize the quadratic equation.
[tex]y^2+6y+5=0\\(y+1)(y+5)=0\\=>y=-1,y=-5[/tex]
The next step is to change back from [tex]y[/tex] to [tex]x[/tex] starting with
[tex]y=-1[/tex].
[tex]x^3=y\\x^3=-1\\=>x=(-1)^{{1}/{3}}=-1[/tex].
The next step is to work out the value of x when [tex]y=-5[/tex].[tex]x^3=y\\x^3=-5\\=>x=(-5)^{{1}/{3}}=-5^{{1}/{3}}[/tex]
