Question

show that (p-1) is factor of (p¹⁰-1) and (p¹¹-1)​

Answer 1

Answer:

Let p−1=0

p=1

For p−1 to be factor, the equation on substituting p=1 should equate to 0

i) f(p)=p  

10

−1

f(1)=1−1=0

ii) f(p)=p  

11

−1

f(1)=1−1=0

∴ Hence p−1 is a factor of both

p  

10

−1 & p  

11

−1

Answer 2

$\huge\sf{\pink{\underline{\underline {♡Answer}}}}$

$\bf f(x) = p - 1 = 0$

$\bf p = 1$

Putting p = 1 in first equation

$\bf \implies {p}^{10} - 1$

$\bf \implies {1}^{10} - 1$

$\bf \implies 1 - 1$

$\bf \implies 0$

Now putting p = 1 in second equation

$\bf \implies {p}^{11} - 1$

$\bf \implies {1}^{11} - 1$

$\bf \implies 1 - 1$

$\bf \implies 0$

Therefore, p -1 is factor of both the equations.