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Re: Both x and y are integers. Which of the following expression [#permalink]
2
Carcass wrote:
Both x and y are integers. Which of the following expressions must be the square of an integer?

Indicate all such expressions.


A. \((x+y)(x-y)+8xy+17y^2\)

B. \(9x^4-12x^2y^2+4y^4\)

C. \(x^6+2x^3y^3+y^6\)


A. Given: \((x+y)(x-y)+8xy+17y^2\)
Expand and simplify the first product: \(x^2-y^2+8xy+17y^2\)
Simplify: \(x^2+8xy+16y^2\)
Factor: \((x + 4y)(x + 4y) = (x + 4y)^2\)
WORKS!


B. Given: \(9x^4-12x^2y^2+4y^4\)
Factor to get: \((3x^2-2y^2)(3x^2-2y^2) = (3x^2-2y^2)^2\)
WORKS!


C. Given: \(x^6+2x^3y^3+y^6\)
Factor to get: \((x^3+y^3)(x^3+y^3) = (x^3+y^3)^2\)[/quote]
WORKS!

Answer: A, B, C

Cheers,
Brent
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Re: Both x and y are integers. Which of the following expression [#permalink]
rishabhrbs96 wrote:
How come A is correct. If we take x=1 and y=1, expression A will be 8 which is not square of any integer.

B is correct because we can factorize it as (3x^2 - 2y^2)^2

C is correct because we can factorize it as (x^3 + y^3)^2

So, the answers should be B and C.


By your own analogy for A ((x+y)(x−y)+8xy+17y^2) becomes 2*0=0+8*1*1=8+17=25, which is 5^2.

Thanks!
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Re: Both x and y are integers. Which of the following expression [#permalink]
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