Determine all pairs of positive integers m n such that 1 xn

Determine all pairs of positive integers (m, n) such that 1 + x^n + x^2n + ... + x^mn is divisible by 1 + x + x^2 + ... + x^m. Solve the equation x^3-3x^2 + 4 = 0, given that two of its roots are equal. Solve the equation x^3 - 9x^2 + 23 x - 15 = 0, given that its roots are in arithmetical progression.

Solution

a)

Let the roots be:a,a,b

SInce we have repeated root a. SO the derivative of the function: g(x)=x^3-3x^2+4 will also have a root x=a

g\'(x)=3x^2-6x

g\'(x)=0 gives

3x^2-6x=0

x^2-2x=0

x=0,2

x=0 is not root of g(x) hence x=2 is the repeated root of g(x)

WE can check

g(2)=2^3-3*2^2+4=0

Product of roots is negative of constant term in g(x) ie -4

SO, a^2b=-4

Hence, b=-1

Hence roots are:-1,2

b)

Roots are:r-d,r,r+d

So, sum of roots =-(-9)=9=3r

Hence, r=3

Product of roots is

r(r^2-d^2)=-(-15)=15

3(9-d^2)=15

9-d^2=5

d^2=4

d=+-2

Sum of roots taken two at a time =23

Case 1. d=2

Roots are: 1,3,5

Sum of roots two at a time =1*3+3*5+5=23

Case 2. d=-2

Roots are again:1,3,5

Hence solution is:x=1,3,5