Question

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Answer 1

If x+y = 1 and x^3 +y^3 = 19 , then find the value of x^2 + y^2

Answer 2

@satellite73

Answer 3

Here is one way to do it


x+y = 1
(x+y)^3 = 1^3
(x+y)^3 = 1
x^3+3x^2y+3xy^2+y^3 = 1
x^3+y^3+3x^2y+3xy^2 = 1
19+3x^2y+3xy^2 = 1
19+3xy(x+y) = 1
3xy(x+y) = 1-19
3xy(x+y) = -18
3xy(1) = -18
3xy = -18
xy = -18/3
xy = -6



x+y = 1
(x+y)^2 = 1^2
(x+y)^2 = 1
x^2+2xy+y^2 = 1
x^2+y^2+2xy = 1
x^2+y^2 = 1-2xy
x^2+y^2 = 1-2(-6)
x^2+y^2 = 1+12
x^2+y^2 = 13

Answer 4

I got the answer

Answer 5

Thank you for helping