Is (x + a + b)^7 - x^7 - a^7 - b^7 Divisible by x^2 + (a + b)x + ab?

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The problem of determining if (x + a + b)^7 - x^7 - a^7 - b^7 is divisible by x^2 + (a + b)x + ab is approached by recognizing that the factors of the divisor are (x + a) and (x + b). To prove divisibility, it is necessary to show that the expression is divisible by both factors. A suggested method involves rewriting terms and simplifying the expression. While the task may seem daunting due to the complexity of expanding a trinomial to the seventh power, it is emphasized that a systematic approach can yield results. Ultimately, the problem is solvable with careful manipulation of the algebraic expressions involved.
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My maths teacher says this problem is not as impossible as it seems, but I just can't solve it.

Show that (x + a + b)^7 - x^7 - a^7 - b^7 is divisble by
x^2 + (a + b)x +ab.
 
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Looking for the easy way out huh ?
You can always solve the entire excercise...

Live long and prosper.
 
Show that (x + a + b)^7 - x^7 - a^7 - b^7 is divisble by x^2 + (a + b)x +ab.

Hint:
Notice that (x+a) and (x+b) are the 2 factors of x^2 + (a + b)x +ab.
So it is equivalent to show that (x + a + b)^7 - x^7 - a^7 - b^7 is divisible by both (x+a) and (x+b).

Let f(x) = (x + a + b)^7 - x^7 - a^7 - b^7
...
...
...
...


Can you continue from here?

Hope this help. :smile:
 
just write everything out
eg. (x+a)^2=x^2+2xa+a^2

maybe rewrite some terms then and you will see that it is divisible by x^2 + (a + b)x +ab
 
KL has the easy way!

Writing it out however... *shudder* I wouldn't wish writing out a trinomial to the 7th power to anyone!

Hurkyl
 
Seemingly by some mathematical coincidence, a hexagon of sides 2,2,7,7, 11, and 11 can be inscribed in a circle of radius 7. The other day I saw a math problem on line, which they said came from a Polish Olympiad, where you compute the length x of the 3rd side which is the same as the radius, so that the sides of length 2,x, and 11 are inscribed on the arc of a semi-circle. The law of cosines applied twice gives the answer for x of exactly 7, but the arithmetic is so complex that the...

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