| Author | Maths Question |
|---|
I cant solve it so need help.
Factorize-
a^3 - 1/a^3 - 2a + 2/a |
| [Post deleted by moderator Brilliant // As requested by author] |
| [Post deleted by moderator Brilliant // As requested by author] |
Sorry, there was a sign mistake in my post #2, a '+' I've mistakenly flipped it into a '-'. I kindly ask the moderators to delete my post #2 and #3.
I'll review the whole argument here:
a^3 - 1/a^3 - 2a + 2/a = 1/a^3 * (a-1) * (a+1) * (a-sqrt(r1)) * (a+sqrt(r1)) * (a-sqrt(r2)) * (a+sqrt(r2))
where r1 and r2 are the constants:
r1 = (1 + i*sqrt(3))/2, and r2 = (1 - i*sqrt(3))/2
Proof:
a^3 - 1/a^3 - 2a + 2/a = a^3 - (1/a)^3 - 2(a-1/a) = (a-1/a)*(a^2 + 1 + (1/a)^2) - 2(a-1/a) = (a-1/a)*(a^2 -1 +(1/a)^2)
Now take aside (a^2 -1 +(1/a)^2), and denote b = a^2:
a^2 -1 +(1/a)^2 = b-1+1/b = 1/b * (b^2-b+1)
Now b^2 -b +1 =0 has (complex) roots r1 = (1+i*sqrt(3))/2 and r2 = (1-i*sqrt(3))/2, hence replacing also b=a^2:
a^2 -1 +(1/a)^2 = (1/a)^2 * (a^2 - r1) * (a^2 - r2) = (1/a)^2 * (a-sqrt(r1)) * (a+sqrt(r1)) * (a-sqrt(r2)) * (a+sqrt(r2))
If you want to get rid of fractionary 1/a inside the first paranthesis, then you can modify it as such:
a-1/a = 1/a * (a^2-1) = 1/a * (a-1) * (a+1)
and so the final result would be:
a^3 - 1/a^3 - 2a + 2/a = 1/a^3 * (a-1) * (a+1) * (a-sqrt(r1)) * (a+sqrt(r1)) * (a-sqrt(r2)) * (a+sqrt(r2))
If you don't know what complex numbers are, and so you don't want to factor this over the complex numbers, but only over the reals, then over the reals b^2-b+1 = a^4 - a^2 + 1 is irreducible, so the final answer would be:
a^3 - 1/a^3 - 2a + 2/a = 1/a^3 * (a-1) * (a+1) * (a^4 - a^2 + 1) |
^^ hmmm i dont understand what u did naviron but i think his task is just to reduce the equation.....
so i would say:
a^3-1/a^3-2a+2/a |*a^3
=a^6-1-2a^4+2a^2
=a^2*(a^4-1-2a^2+2)
=a^2*(a^4-2a^2+1)
now in brackets there is a "binomische formel"-i just know the german word, but i mean the: (a-b)^2=a^2-2ab+b^2 thing...... .
so u can also write:
=a^2*(a^2-1)^2
if there is a mistake or i misunderstood ur task im sorry but school is out now ^^ |
@#5: This is wrong:
=a^6-1-2a^4+2a^2
=a^2*(a^4-1-2a^2+2)
It should be:
a^6 -1 -2a^4 + 2a^2 = a^2 * (a^4 - 1/a^2 - 2a^2+2)
So whatever follows afterwards is incorrect. |
