Finding the roots of a quartic

Question

How do I solve the following equation?

$$(x+3)^5-(x+1)^5=7$$

I tried opening it up, but it turns into an ugly quartic that doesn't factor. I don't know what to do next. Please help me out.

Answer 1

Expanding It as it is will not help much. But before you expand it a small substitution can make it simple We have $$(x+3)^5-(x+1)^5=7$$ Now substitute $ x+2=t $ to get $$(t+1)^5 - (t-1)^5=7$$ $$10t^4+20t^2+2=7$$ Now $p=t^2$ and then you can use quadratic formula to find it's roots and then find out solutions of $x$

Answer 2

You can let $y=x+2$. Then $x+3=y+1$ and $x+1=y-1$. And also you should use the following:

$m^5-n^5=(m-n)(m^4+m^3n+m^2n^2+mn^3+n^4).$

May it helps you!

Answer 3

expanding your equation we get $10\,{x}^{4}+80\,{x}^{3}+260\,{x}^{2}+400\,x+235=0$ and the two real solutions are
$x_1=-2-1/2\,\sqrt {-4+2\,\sqrt {6}}$
$x_2=-2+1/2\,\sqrt {-4+2\,\sqrt {6}}$