Question

if x+y=12and xy=27 find the value of x^3+y^2​

Answer 1

Answer:

$x + y = 12$

• $x = 12 - y$

$xy = 27$

pitting equation of X in above equation,

$(12 - y)y = 27$

$12y - {y}^{2} = 27$

$- {y}^{2} + 12y - 27 = 0$

${y}^{2} - 12y + 27 = 0$

${y}^{2} - 9y - 3y + 27 = 0$

$y(y - 9) + 3(y - 9) = 0$

$(9 - y)(y - 3) = 0$

from above equation we get,

[ y = 3,9 ]

putting y = 3 in equation of X,

$x = 12 - y$

$x = 12 - 3$

$x = 9$

similarly for y = 9 we get X = 3

• points are (x,y) = (3,9)

putting above value in the equation given in the question,

${x}^{3} + {y}^{2}$

${3}^{3} + {9}^{2}$

$27 + 81$

$108$

• for (x,y) = (9,3)

${9}^{3} + {3}^{2}$

$729 + 9$

$738$