The real solutions to the the septic equation \[{x^7} - 3{x^6} - 8{x^5} + 24{x^4} - 7{x^3} + 21{x^2} - 18x + 54 = 0\] are:
(A) 3, 3, -3
(B) -3, -3, 3
(C) No real solutions
(D) Seven overall
Solution
Grouping, $({x^7} - 3{x^6}) + ( - 8{x^5} + 24{x^4}) - (7{x^3} - 21{x^2}) + ( - 18x + 54) = 0$
$ \Rightarrow {x^6}(x - 3) - 8{x^4}(x - 3) - 7{x^2}(x - 3) - 18(x - 3) = 0$
$ \Rightarrow (x - 3)({x^6} - 8{x^4} - 7{x^2} - 18) = 0$
So, x = 3 is one solution.
Now, ${x^6} - 8{x^4} - 7{x^2} - 18 = 0$
Let, ${x^2} = t$
$\therefore {t^3} - 8{t^2} - 7t - 18 = 0$
t = 9 satisfies the equation.
$\therefore {t^2}(t - 9) + t(t - 9) + 2(t - 9) = 0$
$ \Rightarrow (t - 9)({t^2} + t + 2) = 0$
$\therefore t = 9 = {x^2}$, ${t^2} + t + 2 \ne 0$ since ${t^2} + t + 2 > 0$
$\therefore x = \pm 3$
$\therefore x = 3,3, - 3$
Hence, (A).
