## When (x+y) and (x^2 + y^2) are given

$x^3 + y^3 = (x^2 + y^2)(x + y) - 1/2 \; ((x + y)^2 - (x^2 + y^2))(x + y)$
$x^4 + y^4 = (x^3 + y^3)(x + y) - 1/2 \; ((x + y)^2 - (x^2 + y^2))(x^2 + y^2)$
$x^5 + y^5 = (x^4 + y^4)(x + y) - 1/2 \; ((x + y)^2 - (x^2 + y^2))(x^3 + y^3)$
$x^6 + y^6 = (x^5 + y^5)(x + y) - 1/2 \; ((x + y)^2 - (x^2 + y^2))(x^4 + y^4)$
$x^7 + y^7 = (x^6 + y^6)(x + y) - 1/2 \; ((x + y)^2 - (x^2 + y^2))(x^5 + y^5)$
$x^8 + y^8 = (x^7 + y^7)(x + y) - 1/2 \; ((x + y)^2 - (x^2 + y^2))(x^6 + y^6)$
$x^9 + y^9 = (x^8 + y^8)(x + y) - 1/2 \; ((x + y)^2 - (x^2 + y^2))(x^7 + y^7)$
………………..

If   $x + y = a$    and    $x^2 + y^2 = b$,   then

$x^3 + y^3 = 1/2 \; (3 \, a \, b - a^3)$
$x^4 + y^4 = 1/2 \; (-a^4 + 2 \, a^2 \, b + b^2)$
$x^5 + y^5 = 1/4 \; a \, (5 \, b^2 - a^4)$
$x^6 + y^6 = 1/4 \; b \, (-3 \, a^4 + 6 \, a^2 \, b + b^2)$
$x^7 + y^7 = 1/8 \; a \, (a^6 - 7 \, a^4 \, b + 7 \, a^2 \, b^2 + 7 \, b^3)$
$x^8 + y^8 = 1/8 \; (a^8 - 4 \, a^6 \, b - 2 \, a^4 \, b^2 + 12 \, a^2 \, b^3 + b^4)$
$x^9 + y^9 = 1/16 \; a \, (a^8 - 18 \, a^4 \, b^2 + 24 \, a^2 \, b^3 + 9 \, b^4)$