Show that $\frac p{a^n}-\frac q{b^n}-1=0$.

Question

If $x^n-py^n-qz^n$ is exactly divisible by $x^2-(ay+bz)x+abyz$ , then prove that $\frac p{a^n}-\frac q{b^n}-1=0$

@DanZimm You may try to use factor theorem for solving this problem. — Snehil Sinha, Jun 17 '14 at 08:56
Are u sure it is $p/a^n-q/b^n=1$. I think it should be $p/a^n+q/b^n=1$. — Samrat Mukhopadhyay, Jun 17 '14 at 09:07
@SamratMukhopadhyay no i am sure that my question is correct.But if you think it should be $p/a^n+q/b^n=1$ then please send your answer according to it. — Snehil Sinha, Jun 17 '14 at 09:10

score 3 · Accepted Answer · answered Jun 17 '14 at 09:10

3

Erom the second quadratic we can see that the roots are $$\alpha=ay,\beta=bz$$ which implies that they are roots of $x$ for the first equation also. So, we get the equations $$a^ny^n=py^n+qz^n\\ b^nz^n=py^n+qz^n$$ Hence, $$ a^ny^n=py^n+q(ay)^n/b^n\Rightarrow p/a^n+q/b^n=1$$

answered Jun 17 '14 at 09:10

Samrat Mukhopadhyay

16,277

Mukhopa Can you plz explain how did u derive $a^ny^n=py^n+q(ay)^n/b^n⇒p/a^n+q/b^n=1 $ – Snehil Sinha Jun 19 '14 at 06:49
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Just divide both sides by $(ay)^n$ – Samrat Mukhopadhyay Jun 19 '14 at 11:56
thanx for help I got it – Snehil Sinha Jun 20 '14 at 06:42

Show that $\frac p{a^n}-\frac q{b^n}-1=0$.

1 Answers1