WilDThiNg wrote:

MathRevolution wrote:

==>If you substitute x=1, from 1^6+1=(1+n+1)(1-n+1), you get 2=(2+n)(2-n)=4-n^2. Naturally, you get n^2=4-2=2, and n=±√2. Therefore the answer is C.

Answer: C

Hi, please can you show how to solve it alegebraically?

Thanks

Solving the Right hand side which is \((x^3+nx+1)(x^3-nx+1) = (x^3+1)^2 - (nx)^2\) ---using the formula \((a+b)(a-b)= a^2-b^2\)

equal to , \(x^6+1+2x^3-n^2x^2\)

equating LHS=RHS

\(x^6+1=x^6+1+2x^3-n^2x^2\)

gives \(2x^3-n^2x^2=0\)

therefore \(n^2=2x\)

n=\(±\sqrt{2x}\)

When x=0, n=0

when x=1, n=\(±\sqrt{2}\) which is option C

If u found my post useful, press kudos!
_________________

Please give kudos, if you like my post

When the going gets tough, the tough gets going...