#2**+10 **

First, let us show that it is true for n=1

So....a^(2(1) + 1) + b^(2(1) + 1) = a^3 + b^3 = (a +b)(a^2 -ab + b^2)...and (a + b) is a factor

Now, let us assume it is true for k=1, that is, a^(2(k) + 1) + b^(2(k) + 1) is true

Now, let's prove that it is true for k + 1

So we have

a^(2(k+1)+ 1) + b^(2(k+1) + 1) =

a^(2k + 3) + b^(2k + 3) =

a^3*a^(2k) + b^3*b^(2k)......and using a^3 = (a +b)(a^2 -ab + b^2) - b^3 , we can write

[(a+b)(a^2 -ab + b^2) - b^3]a^(2k) + b^3*b^(2k) =

(a+b)(a^2 -ab + b^2)a^(2k) - b^3[ a^(2k) - b^(2k)]

Notice that the last term is a difference of two even powers.........and it can be shown that the difference of two even powers can be factored with (a + b) as a factor thusly...

(a^(2n) - b^(2n)) = (a + b)[a^(2n-1) - a^(2n-2)b + a^(2n-3)b^2 - a^(2n-4)b^3 +.....+ ab^(2n-2) - b^(2n-1) ]

Thus, (a+b) is a factor of both terms, so (a +b) is a factor of a^(2k + 3) + b^(2k + 3)

CPhill Jan 5, 2015

#1

#2**+10 **

Best Answer

First, let us show that it is true for n=1

So....a^(2(1) + 1) + b^(2(1) + 1) = a^3 + b^3 = (a +b)(a^2 -ab + b^2)...and (a + b) is a factor

Now, let us assume it is true for k=1, that is, a^(2(k) + 1) + b^(2(k) + 1) is true

Now, let's prove that it is true for k + 1

So we have

a^(2(k+1)+ 1) + b^(2(k+1) + 1) =

a^(2k + 3) + b^(2k + 3) =

a^3*a^(2k) + b^3*b^(2k)......and using a^3 = (a +b)(a^2 -ab + b^2) - b^3 , we can write

[(a+b)(a^2 -ab + b^2) - b^3]a^(2k) + b^3*b^(2k) =

(a+b)(a^2 -ab + b^2)a^(2k) - b^3[ a^(2k) - b^(2k)]

Notice that the last term is a difference of two even powers.........and it can be shown that the difference of two even powers can be factored with (a + b) as a factor thusly...

(a^(2n) - b^(2n)) = (a + b)[a^(2n-1) - a^(2n-2)b + a^(2n-3)b^2 - a^(2n-4)b^3 +.....+ ab^(2n-2) - b^(2n-1) ]

Thus, (a+b) is a factor of both terms, so (a +b) is a factor of a^(2k + 3) + b^(2k + 3)

CPhill Jan 5, 2015