How can it be factorised?
Find the sum of all real k such that x + y + z divides x 3 + y 3 + z 3 + k x y z for all x , y , z .
The answer is -3.0.
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1 solution
@Chew-Seong Cheong Sir, why did you check only -3.......How do you know that there aren't any other values.??
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I did not check. It was derived from the equation.
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No Sir, that is visible.......What I am asking is that if k had some other value (not necessarily an integer ) then, how would we know that no solutions exist then........???
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@Aaghaz Mahajan – I don't think there are other values.
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@Chew-Seong Cheong – But Sir, is there any proof????
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@Aaghaz Mahajan – My solution is the proof. The solution is for all x , y and z .
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@Chew-Seong Cheong – Right Sir........Thank you!!
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@Aaghaz Mahajan – To answer your question you can either identify the elementary formula right away, but true that doesn't tell you if there exists any other k such that the questions hold. For that purpose use the remainder theorem, and make the equation identically 0 . and it follows that k has to be 3 .
For all x , y and z , we have x 3 + y 3 + z 3 = ( x + y + z ) ( x 2 + y 2 + z 2 − x y − y z − z x ) + 3 x y z . Therefore, if k = 0 or k = − 3 , x 3 + y 3 + z 3 + k x y z = ( x + y + z ) ( x 2 + y 2 + z 2 − x y − y z − z x ) which is divisible by x + y + z . Therefore the sum of all valid real k is 0 − 3 = − 3 .