If $$x^n-py^n+qz^n$$ is divisible by $$x^2+abyz-bzx-axy$$ , then what is $$\frac{p}{a^n}-\frac{q}{b^n}$$ equal to?

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)