Question

Given x+y=2 & xy=3, evaluate $x^5+y^5$.

Solution

Putting y=2-x in the equation xy=3,

x(2-x)=3

Or, x²-2x+3=0

Or, $x=\frac {2\pm\sqrt {4-12}}{2}$

Or, $x=1\pm i.\sqrt 2$

Out of these two complex conjugate roots, one should represent x and the other should represent y.

Let $x=1+ i.\sqrt 2$, So $y=1- i.\sqrt 2$

${x^5} + {y^5} = {(1 + i\sqrt 2 )^5} + {(1 – i\sqrt 2 )^5}$

$ = [{}^5{C_0} + {}^5{C_1}i\sqrt 2 + {}^5{C_2}{(i\sqrt 2 )^2} + {}^5{C_3}{(i\sqrt 2 )^3} + {}^5{C_4}{(i\sqrt 2 )^4} + {}^5{C_5}{(i\sqrt 2 )^5}]$$+$$[{}^5{C_0} – {}^5{C_1}i\sqrt 2 + {}^5{C_2}{(i\sqrt 2 )^2} – {}^5{C_3}{(i\sqrt 2 )^3} + {}^5{C_4}{(i\sqrt 2 )^4} – {}^5{C_5}{(i\sqrt 2 )^5}]$

$= 2[{}^5{C_0} + {}^5{C_2}{(i\sqrt 2 )^2} + {}^5{C_4}{(i\sqrt 2 )^4}]$

$ = 2\left[ {1 + \frac{{5!}}{{2!3!}}( – 1)(2) + 5(4)} \right]$

=2(1-20+20)

=2