how do you factor sums of cubes;t^6+1

Question

how do you factor sums of cubes like t⁶ +1 thank you in advance

Answer 1

f(t) = t^6 + 1 = ((t^2)^3 + 1^3)

f(t) = (t^2 + 1)((t^2)^2 - t^2*1 + 1^2)

 

Which is what Vivian found. But you can go farther:

f(t) = (t^2 - i^2)(t^4 - t^2 + 1)

If t^4 - t^2 + 1 = 0,

we can use Quadratic Formula:

t^2 = (1 ± √(1-4))/2 = (1 ± i√(3))/2

t = ± √( ( 1 ± i√(3) )/2 )

So fully factored:

f(t) = (t ± i)(t ± √( ( 1 ± i√(3) )/2 ))

All of the 6 zeros are imaginary; because t^6 + 1 is always positive.

Answer 2

Hi Sammy;

t⁶+1

[(t²)³]+1

Let's say that x=t²

x³+1

According to Purple Math

http://www.purplemath.com/modules/specfact2.htm

x³ + y³ = (x + y)(x² – xy + y²)

Input our information...

y³=1³=1

xy=(x)(1)=1x=x

y²=1²=1

x³+y³=x³+1=(x+1)(x²-x+1)

Let's replace x with t²...

(t²+1){[(t²)²]-t²+1)}

(t²+1)(t⁴-t²+1)