Can someone please factorise this expression (its in mathematics)?

Question

Note: If the number is after the variable it is an exponent if it is before the variable it is a coefficient

The expression is a3 - 1/a3 -2a + 2/a
Thank You

Answer 1

The standard formula for factorising difference of two cubes

x^3 - y^3 = (x-y)(x^2+xy+y^2)
gives us factors of the first pair of terms
(a - 1/a)(a^2 + 1 + 1/(a^2))

The second pair is -2(a - 1/a)

Now these two expressions have a common factor, a - 1/a, so taking out this common factor gives us

(a - 1/a)(a^2 + 1 + 1/(a^2) -2)
= (a - 1/a)(a^2 - 1 + 1/(a^2))

This is a much later PS (40 minutes later, in fact). I'm pleased to see avunger agrees with my solution. The reason I've come back is that it occurs to me that we can go further with the second factor.

If we write it as a^2 + 2 + 1/(a^2) - 3, then it becomes

(a + 1/a)^2 -3
= (a + 1/a + √3)(a + 1/a - √3), using the standard "difference of squares" formula, but we don't usually use irrational numbers in these sorts of factors.

I notice Helmut has √3 in his solution, but it also includes i so I expect (without checking his work) that he's made a mistake somewhere.

Answer 2

a^3 - 1/a^3 - 2a + 2/a =
(a^6 - 1 - 2a^4 + 2a^2)/a^3 =
(a^6 - 2a^4 + 2a^2 - 1)/a^3 =
(a^2 - 1)(a^4 - a^2 + 1)/a^3 =
(a + 1)(a - 1)(a^4 - a^2 + 1/4 -1/4 + 1)/a^3 =
(a + 1)(a - 1)((a^2 - 1/2)^2 + 3/4 )/a^3 =
(a + 1)(a - 1)((a^2 - 1/2)^2 - (- 3/4 )/a^3 =
(a + 1)(a - 1)((a^2 - 1/2 + (i/2)√3)(a^2 - 1/2 - (i/2)√3))/a^3 =
(a + 1)(a - 1)((a^2 - 1/2(1 + i√3)) (a^2 - 1/2(1 - i√3)) /a^3 =
(a + 1)(a - 1)((a + √(1/2(1 + i√3)))(a - √(1/2(1 + i√3))) (a + √(1/2(1 - i√3))) (a - √(1/2(1 - i√3)))/a^3

Answer 3

Hopefully, you purposely didn't parenthesize, because right now I'm seeing ambiguity as to what this question can be. Assuming it's this:

a^3 - 1/(a^3 - 2a) + 2/a

We factor the denominator of the second term to get

a^3 - 1/[a(a^2 - 2)] + 2/a

Now, we put them all under a common denominator of a(a^2 - 2).

[ a^3[a(a^2 - 2)] - 1 + 2(a^2 - 2) ] / [a(a^2 - 2)]

Simplifying,

[ a^4[a^2 - 2] - 1 + 2a^2 - 4 ] / [a(a^2 - 2)]

And I think I misread your question because a power of 6 is involved so I'll stop here.

Answer 4

a3-1/a3-2a+2/a
as a3 - b3 can be written as (a - b)(a2+ab+b2)

a3-1/a3 = (a - 1/a)(a2+a*1/a+1/a2)
=(a - 1/a)(a2+1+1/a2)

and in -2a + 2/a,
-2 can be taken as common

-2a+ 2/a = -2(a - 1/a)

so the equation is
(a -1/a)(a2 + 1 + 1/a2) - 2(a - 1/a)

taking (a - 1/a) as common
(a - 1/a)(1(a2 + 1 + 1/a2) - 2(1))
= (a - 1/a)(a2 +1 + 1/a2 -2)
=(a - 1/a)(a2 - 1 + 1/a2)

Answer 5

a^3 - 1/a^3 - 2a + 2/a
a^3 - a^(-3) - 2a + 2a^(-1)
a^(3 -(-3) ) - 2a^(1 + (-1) )
a^6 - 2a^0
a^6 - 2 . 1
a^6 - 2

Answer 6

21a^2b^2-2ab in simple terms aspect out the elementary aspect... then ab(21ab-2) ... interior the subsequent equation.,,,, 2ax+3ay-4bx+6by then a(2x+3y) - 2b(2x+3y) then.. (a-2b)(2x+3y) in simple terms aspect out the elementary aspect again,,, sorry for the edit,, i didnt see that the in simple terms proper time period is -6by desire it ought to help... tnx :D