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How to find the value of n in the equation n(n^2-1)=990?

Please explain the derivation along with the answer.
Yes but how do we get it.PLease explain the method.
I hope, I cleared it, right


Write n(n+1)(n-1)=990=11*9*10
See that (n-1)(n)(n+1) are 3 consequtive integers. and the prome factorization of 990 yeilds 9*10*11  Which is what we want, So, n=10
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actually, We cannot find all roots of equation by hand, If it was possible, we could have solutions to all polynomials. The way of finding roots to all equations is not possible, by hand,. U surely need calculator to do these ques. But there are some methods through which u can find solutions to some special polynomials, like differentiating a function will give anther function, and if our original function has repeated roots, then, that repeated root will also be a root of its derivative,.
Also veita's method, Veita root jumpind are some other ways, For this ques, u need to use calculator
What u can get, without using calculator is that one root is positive and other is negative(Do u want me to explain this), There is also a way through which u can ger an approximated root, But with help of calculus, So, I wont mention here
Thank you very much .......
How to find the value of n in the equation n(n^2-1)=990?
Please explain the derivation along with the answer.
             SORRY I Cannot explain the solution , but by hit and trial method , the answer is 10 .

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I gusee not, because, it yeilds a easy form, See my solution :)
yaar your solution is good , but hit and trial method is much easier than that of yours .. Thanks for ur solution :)
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