Squares and Cubes
I. x 2 = y 2 ⇒ x = y II. x 3 = y 3 ⇒ x = y
Which of the above statements is/are true for real x and y ?
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3 solutions
Correction, statement 2 might be true because cube-function doesn't alter the sign of the input. You still need to prove that there are no such real numbers x = y where x 3 = y 3 .
Statement 1 x 2 = y 2 ⇒ ( x − y ) ( x + y ) = 0
⇒ ( x − y ) = 0 or x = y ⇒ ( x + y ) = 0 or x = − y
Statement 2
x 3 = y 3 ⇒ x 3 − y 3 = 0
⇒ ( x − y ) ( x 2 + x y + y 2 ) = 0
⇒ ( x − y ) = 0 or x = y
setment 2: where is the (x^2+xy+y^2) ?????? I don't understand your solution........
You need to prove that x 2 + x y + y 2 = 0 for real x and y .
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try putting it through the quadratic formula and observe the value under the root, it should be negative, if not reply to me
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Yes, that's one way to prove it, but the answer is still just assuming that non-obvious fact without a proof which is probably what I meant.
Statement 1 lacks the ±.
It should have been x 2 = y 2 ⇒ ± x = ± y
Always remember the ± when dealing with even powers.
Statement 1 is false because we could have x = 1 and y = − 1 . While x 2 = y 2 , it does not imply that x = y .
However, statement 2 is true because the cube root function does not return a ± . x 3 = y 3 ⟹ x = y