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N 2 N^{2} + N N = N 3 N ^ {3} - N N

What is the product of all the numbers N could be?


The answer is 0.

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4 solutions

Siva Budaraju
Sep 10, 2017

The easiest solution to find is N=0, by moving all the terms to one side and factoring an N out. Now we do not need to find the other solutions because 0 times anything is 0 \huge 0 .

Munem Shahriar
Dec 8, 2017

N = 0 N = 0 is a solution. Hence the product must be 0.

Sumukh Bansal
Nov 19, 2017

There are three solutions of the equation N 2 N^{2} + N N = N 3 N ^ {3} - N N namely ( 2 , 1 , 0 ) (2,-1,0) and 2 × ( 1 ) × ( 0 ) = 0 2\times(-1)\times(0)=0

There are three solutions to the above equation: 2, -1, and 0. Therefore, their product is 2 0 1 2*0*-1 , which is equal to 0 \boxed{0} .

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