Please explain how to solve this16 For all positive integers n 1, {x(xn-1 - Maths - Principle of Mathematical Induction

Question

P l e a s e   e x p l a i n   h o w   t o   s o
l v e   t h i s 16   F o r   a l l   p o s i t
i v e   i n t e g e r s   n > 1 ,      
        { x ( x n - 1   -   n a n - 1
)   +   a n ( n - 1 ) }   i s   d i v i s i b l
e   b y              
      1 )   ( x - a ) 2      
  2 )   x - a       3 )   2 ( x - a )
          4 )   x + a

Ghanshyam Dhakar · Accepted Answer

Dear student,
xxn-1-nan-1+ann-1     
eq1=xn-xnan-1+nan-an=xn-an-xnan-1+nanuse xn-an = x-axn-1+xn-2a+
xan-2+an-1=x-axn-1+xn-2a+ xan-2+an-1-nan-1x-a=x-axn-1+xn-2a+
xan-2+an-1-nan-1=x-axn-1-an-1+xn-2a-an-1+
+xan-2-an-1+an-1-an-1=x-axn-1-an-1+xn-2a-an-1+
+xan-2-an-1+an-1-an-1=x-axn-1-an-1+axn-2-an-2+
+an-2x-a+0use xn-1-an-1 = x-axn-2+xn-3a+
xan-3+an-2   and xn-2-an-2 = x-axn-3+xn-4a+
xan-4+an-3  =x-ax-axn-2+xn-3a+ xan-3+an-2+ax-axn-3+xn-4a+
xan-4+an-3+ +an-2x-a=x-a2xn-2+xn-3a+ xan-3+an-2+axn-3+xn-4a+
xan-4+an-3+
+an-2eq1 is divisible by x-a2    answer
Regards

P l e a s e e x p l a i n h o w t o s o l v e t h i s 16 . F o r a l l p o s i t i v e i n t e g e r s n > 1 , { x ( x n - 1 - n . a n - 1 ) + a n ( n - 1 ) } i s d i v i s i b l e b y 1 ) ( x - a ) 2 2 ) x - a 3 ) 2 ( x - a ) 4 ) x + a

$P l e a s e e x p l a i n h o w t o s o l v e t h i s 16 . F o r a l l p o s i t i v e i n t e g e r s n > 1, {x (x^{n - 1} - n . a^{n - 1}) + a^{n} (n - 1)} i s d i v i s i b l e b y 1) (x - a)^{2} 2) x - a 3) 2 (x - a) 4) x + a$