Algebra problem?

Question

If: x+y=a , x^2 + y^2=b and x^3 + y^3= c
prove that:
a^3=3ab-2c

Answer 1

We have x+y=a,x^2+y^2=b and x^3+y^3=c
Therefore 2xy=(x+y)^2-(x^2+y^2)
=a^2-b
Therefore xy=(a^2-b)/2and 3xy={3(a^2-b)/2
=(3a^2-3b)/2
a^3=(x+y)^3
=x^3+y^3+3xy(x+y)
=c+{(3a^2-3b)/2}*a
=c+(3a^3-3ab)/2
=(2c+3a^3-3ab)/2
Multiplying both sides by 2 we get,
2a^3=2c+3a^3-3ab
=>2a^3-3a^3=2c-3ab
=> -a^3=2c-3ab
=>a^3=3ab-2c

Answer 2

Substitute a for x + y, b for x^2 + y^2 and c for x^3 + y^3
(x+y)³ = 3(x+y)(x²+y²) - 2(x³ + y³)

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Left part:
Expand (x+y)^3:
x³ + 3x²y + 3xy² + y³
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Right part:

Expand 3(x+y)(x²+y²)
3(x³ + x²y + xy² + y³) =
3x³ + 3x²y + 3xy² + 3y³

and then subtract (2x³ + 2y³):
3x³ + 3x²y + 3xy² + 3y³ - 2x³ - 2y³ =
x³ + 3x²y + 3xy² + y³ ... (because 3x³ - 2x³ = x³ and 3y³ - 2y³ = y³)
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Left = Right
x³ + 3x²y + 3xy² + y³ = x³ + 3x²y + 3xy² + y³

Answer 3

a^2=x^2+y^2+2xy
a^2=b+2xy

a^2-b=2xy
xy=(a^2-b)/2


x^3 + y^3= c
(x+y)(x^2+y^2-xy)=c
a.(b-xy)=c
a(b-(a^2-b)/2)=c
2ab-a^3+ab=2c
a^3=3ab-2c