Asked by Crow
Suppose x and y are real numbers satisfying
x(x+i)+y(y+i)−6(8+i)=0.
What is the value of x^3+y^3?
x(x+i)+y(y+i)−6(8+i)=0.
What is the value of x^3+y^3?
Answers
Answered by
Steve
x(x+i)+y(y+i)−6(8+i)=0
x^2 + ix + y^2 + iy = 48+6i
(x^2+y^2) + (x+y)i = 48+6i
x+y = 6
x^2+y^2 = 48
x=3+√15
y=3-√15
x^3+y^3 = (3+√15)^3 + (3-√15)^3
The odd powers of √15 vanish, leaving us with
2(3^3 + 3(3)(15)) = 324
x^2 + ix + y^2 + iy = 48+6i
(x^2+y^2) + (x+y)i = 48+6i
x+y = 6
x^2+y^2 = 48
x=3+√15
y=3-√15
x^3+y^3 = (3+√15)^3 + (3-√15)^3
The odd powers of √15 vanish, leaving us with
2(3^3 + 3(3)(15)) = 324
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