Solve simultaneously:-
x + y = 5, x^3 + y^3 = 35.
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x^3 + y^3 = (x + y)(x^2 + xy + y^2) = 35.
= 5 { (x+y)^2 - 3xy } =35
5^2 - 3xy = 7
3xy = 18
Xy = 6 =2×3
X= 2
Y = 3
= 5 { (x+y)^2 - 3xy } =35
5^2 - 3xy = 7
3xy = 18
Xy = 6 =2×3
X= 2
Y = 3
Kodkani:
X can be 3 and Y can be 2.
X and Y can also be 1 and 6.
Why only 2 and 3?
First condition given x+y = 5
If we took 1 and 6 .. 1+6=7 .. not satisfied our first equation
How do you know that x has to be 2 and Y has to be 3?
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