if a,b,c,d,e and f are six real numbers such that `a+b+c=d+e+f` `a^2+b^2+c^2=d^2+e^2+f^2` and `a^3+b^3+c^3=d^3+e^3+f^3` , prove by mathematical induction that `a^n+b^n+c^n=d^n+e^n+f^n forall n in N`.

Explanation

Let `P(n):a^(n)+b^(2)+c^(n)=d^(n)+e^(n)+f^(n),AA n in N " " ..(i)`
where `a+b+c+d=e+" "...(ii)`
`a^(2)+b^(2)+c=d^(2)+e^(2)+f " "....(iii)`
and `a^(2)+b^(3)+c^(3)=d^(3)+e^(2)+^(3)" "...(iv)`
Step I from n from Eq. (i) we get
`P(1): a+b+c=d+e+f" " ` [ given]
Hence the result is true for n 1
Also, for n=2 from Eq(i), we get
`P(2): a^(2)+b^(2)+c^(2)=d^(2)+e^(3)+f^(3) " "` [ given]
Hennce the result true or n=3
Therefore, P(1) , P(2) and P (3) are true.
Step II Assume that `P(k-2),P(k-1)and P(k)` are true, then
`P(k-2), a^(k-2)+b^(k-2)= d^(k-2)+e^(k-2)+f^)k-2) " "...(v)`
`p(k-1):a^(-1)+b^(k-1)+c^(k-1)=d^(k-1)+e^(k-1)+f^(k-1) " "....(vi)`
and `P(k): a^(k)+b^(k)+c^(k)=d^(k)+e^(k)+f^(k) " m "...(vii)`
Step III for ` xn=k+1` we shall to prove that
`P(k+1):a^(k+1)+b^(k+1)=d^(k+1)+e^(k+1)+f^(k+1)`
LHS `=a^(k+1)+b^(k+1)+c^(k+1)`
`=(a^(k)+b^(k)(a+b+c)-(a^(k-1)+b^(k-1)+c^(k-1))`
`(ab+bc+ca)+abc(a^(k-2)+b^(k-2)+b^(k-2)+c^(k-2))`
`=(d^(k)+e^(f)+f^(k))(d+e+f)-(d^(k-1)+e^(k-1)+c^(k-2))`
`(de+ef+fd)+def(d^(k-2)+e^(k-2)+f^(k-2))`
[ using Eqs. (ii), (iii), (iv), (v), (vi), (vii)]
`:. (a+b+c)^(2)=(d+e+f)^(2)`
`rArr a^(2)+b^(2)+c^(2)+2(ab+bc+ca)`
`=d^(2)+e^(2)+f^(2)+2(de+ef+fd)`
`rArr ab+bc+ca=de+ef+fd`
`[ :. a^(2)+b^(2)+c^(2)=d^(2)+e^(2)+f^(2)]`
and `a^(3)+b^(3)+c^(3) -3abc`
`(=d+e+f)(d^(2)+e^(2)+f^(2)-de-ef-fd)`
`=d^(3)+e^(3)+f^(3)-edf`
`rArr abc=def [ :. a^(3)+b^(3)+c^(3)=d^(3)+e^(3)+f^(3))`
`=d^(k+1)+e^(k+1)+f^(k-1)=RHS`
This shows that result is true for n=k+1. Hence by second perincipal of mathmatical inducition, the result is true for all `n in N`.