Question

if x+y=12,and xy=27,findx³+y³.


with solution ​

Answer 1

Answer:

$(x + y) {}^{ 3} = {x}^{3} + {y}^{3} + 3xy(x + y) \\ \\ x + y = 12 \\ (x + y) {}^{3} = 12 {}^{3} \\ {x}^{3} + {y}^{3} + 3xy(x + y) = 1728 \\ {x}^{3} + y {}^{3} = 1728 - 3xy(x + y) \\ {x}^{3} + y {}^{3} = 1728 - 3 \times 27(12) = 1728 - 972 \\ {x}^{3} + y {}^{3} = 756$

Answer 2

Answer:

270

Step-by-step explanation:

x+y=12, xy=27,x^3+y^3=?

x+y=12

x=12-y-----------1

xy=27

(12-y)y=27

12y-y^2=27

y^2-12y+27=0

y^2-9y-3y+27=0

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

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

y=9 or y=3

if y=9 then x=3

if y=3 then x=9

x^3+y^3=(3)^3+(9)^3

=27+243

=270