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Re: If x and y are positive integers, and if x^2 + 2xy + y^2 = 25, then wh
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23 May 2023, 04:51
To find the value of (x+y)^3, we can expand the cube of the binomial (x+y) using the binomial theorem.
(x+y)^3 = x^3 + 3x^2y + 3xy^2 + y^3
Given the equation x^2 + 2xy + y^2 = 25, we can see that it is similar to the form (x+y)^2 = 25. Comparing this with the expanded form of (x+y)^3, we can observe that:
x^3 + 3x^2y + 3xy^2 + y^3 = (x^2 + 2xy + y^2)(x+y)
Substituting the given equation, we have:
(x^2 + 2xy + y^2)(x+y) = 25(x+y)
Since x and y are positive integers, we can divide both sides by (x+y) (which is non-zero) to obtain:
x^2 + 2xy + y^2 = 25
This means that (x^2 + 2xy + y^2)(x+y) = 25(x+y)
Substituting this into the expanded form of (x+y)^3:
(x+y)^3 = (x^2 + 2xy + y^2)(x+y) = 25(x+y)
Now we can substitute the value of x^2 + 2xy + y^2 = 25:
(x+y)^3 = 25(x+y)
Since we know that x and y are positive integers, (x+y) must also be positive. Therefore, we can divide both sides of the equation by (x+y):
(x+y)^2 = 25
Taking the square root of both sides, we get:
x+y = 5
Substituting this value back into (x+y)^3:
(x+y)^3 = 5^3 = 125
Therefore, the value of (x+y)^3 is 125.
The correct answer is E. 125.