Answer
Reducing $$x^2+abyz-bzx-axy$$ to the product of two linear factors:
=>$$x^2+abyz-bzx-axy$$
=>$$x^2-bzx-axy+abyz$$
=>$$x(x-bz)-ay(x-bz)$$
=>$$(x-ay)(x-bz)$$
we get x=ay and x=bz as a factor of $$x^2+abyz-bzx-axy$$
Now, since $$x^2+abyz-bzx-axy$$ is a factor of $$x^n-py^n+qz^n$$
∴$$ x^n-py^n+qz^n$$ =(x-ay)(x-bz)Q +0, where Q is quotient.
Now, taking LHS,
$$x^n-py^n+qz^n$$
Put x=ay
=> $$a^ny^n-py^n+qz^n=0 $$ —i)
Put x=bz
=> $$b^nz^n-py^n+qz^n=0 $$ —ii)
eq (ii)-eq (i)
We get, $$a^ny^n=b^nz^n$$
Now,
From eq (ii),$$b^nz^n-py^n+qz^n$$=0
Divide by $$b^nz^n, 1-\frac{py^n}{b^nz^n}+\frac{q}{b^n}=0$$ –iii)
Or, $$1-\frac{p}{a^n}+\frac{qz^n}{a^ny^n}=0$$ —iv) (‘.’ $$a^ny^n=b^nz^n$$)
From (iii) and (iv) we get $$\frac{p}{a^n}-\frac{q}{b^n}$$=1 , Ans
Answer – (C)