Question

How do you factor `64t^6-1`?

Answer 1

`(2t-1)(2t+1)(16t^4+4t^2+1)`

Explanation:

This is a `color(blue)"difference of cubes"` and in general, factorises as follows.

`color(red)(bar(ul(|color(white)(a/a)color(black)(a^3-b^3=(a-b)(a^2+ab+b^2))color(white)(a/a)|)))...... (A)`

Note that.

`(4t^2)^3=64t^6" and " (1)^3=1`

`rArra=4t^2" and " b=1`

substitute these values for a and b into (A)

`(4t^2-1)(16t^4+4t^2+1)`

We now have a factor, `(4t^2-1)` which is a `color(blue)"difference of squares"` and factorises, in general as.

`color(red)(bar(ul(|color(white)(a/a)color(black)(a^2-b^2=(a-b)(a+b))color(white)(a/a)|)))...... (B)`

`(2t)^2=4t^2" and " (1)^2=1`

`rArra=2t" and " b=1`

substitute these values for a and b into (B)

`(2t-1)(2t+1)`

Pulling all this together.

`64t^6-1=(4t^2-1)(16t^4+4t^2+1)`

`=(2t-1)(2t+1)(16t^4+4t^2+1)`

The factor `16t^4+4t^2+1" may be factorised as follows"`

This is a polynomial of degree 4, with no odd degree terms.

`a^4+(2-k^2)a^2b^2+b^4=(a^2-kab+b^2)(a^2+kab+b^2)`

here a = 2t , b = 1 and k = 1

`rArr16t^4+4t^2+1=(4t^2-2t+1)(4t^2+2t+1)`

The whole expression now factorises to.

`64t^6-1=(2t-1)(2t+1)(4t^2-2t+1)(4t^2+2t+1)`

I thank George C, for his input to the factorising of the quartic.

Answer 2

Use the difference of squares, difference of cubes and sum of cubes identities to find:

`64t^6-1 = (2t-1)(4t^2+2t+1)(2t+1)(4t^2-2t+1)`

Explanation:

The difference of squares identity can be written:

`a^2-b^2=(a-b)(a+b)`

The difference of cubes identity can be written:

`a^3-b^3 = (a-b)(a^2+ab+b^2)`

The sum of cubes identity can be written:

`a^3+b^3=(a+b)(a^2-ab+b^2)`

Hence we find:

`64t^6-1 = (8t^3)^2-1^2`

`color(white)(64t^6-1) = (8t^3-1)(8t^3+1)`

`color(white)(64t^6-1) = ((2t)^3-1^3)((2t)^3+1^3)`

`color(white)(64t^6-1) = (2t-1)((2t)^2+(2t)+1)((2t)+1)((2t)^2-(2t)+1)`

`color(white)(64t^6-1) = (2t-1)(4t^2+2t+1)(2t+1)(4t^2-2t+1)`