Question

Solve for $x:$
$(x-1{)}^3+(x-2{)}^3+(x-3{)}^3+(x-4{)}^3+(x-5{)}^3=0$

Answer 1

Observe that A.M.of pair $(x-1),(x-5)$ and $(x-2),(x-4)$ is $(x-3)$
Hence we put $x-3=y$ and change the equation in y.
$\therefore x=y+3.$ Putting for x the given equation becomes
${(y+2)}^3+{(y+1)}^3+y^3+{(y-1)}^3+{(y-2)}^3=0$
$2[y^3+3y(4\left)\right]+2[y^3+3y(1{)}^2]+y^3=0$
or $5y^3+30y=ory(y^2+6)=0$
$\therefore y=0,\pm i\sqrt{6}$ $\therefore x=y+3=3,3\pm i\sqrt{6}.$