# Factorize x^6 + 6x^3 + 5.

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### 3 Answers

We have to factorize x^6 + 6x^3 + 5

First, see that x^6 + 6x^3 + 5 is of the form (x^3)^2 + 6x^3 + 5

Now if we take x^3 as y this is equal to y^2 + 6y + 5

Now 6y can be written as 5y + y

=> y^2 + 5y + y + 5

=> y( y+5) +1(y+5)

=> (y+1)(y+5)

substitute back x^3 for y

=> (x^3 + 1)(x^3 + 5)

Now x^3 +1

=> x^3 + x^2 + x –x^2 – x +1

=> x^3 + x^2 - x^2 – x + x +1

=> x^2(x+1) – x(x+1) + 1(x+1)

=> (x^2 – x +1)(x+1)

Therefore x^6 + 6x^3 + 5

= (x^2 + 5) (x^2 – x +1)(x+1)

**The required factorization of x^6 + 6x^3 + 5 is (x^2 + 5) (x^2 – x +1)(x+1)**

The task is to

Factorize x^6 + 6x^3 + 5

My first motivation was to see that x^6 is related to x^3 by a power of 2.

Hence I take the liberty to let y=x^3, so that x^6 would become y^2.

The equation now becomes a quadratic equation in y:

y^2 + 6y + 5

We can now do our usual to factorize this quadratic equation to become:

(y+5)(y+1)

If we replace y by x^3 again, we would then get:

** (x^3 +5).(x^3+1)**

The first factor (x^3+5) can remain as it is, since 5 is not a cubic number.

However, the 2nd factor (x^3+1) is a little elusive. As we know, the number 1 raised to any power still yields 1. Hence, this 1 is a cubic number.

Also, we know that there are missing terms in the polynomial: x^2 and x terms. We cater for that by deliberately including the adding and subtracting of the same term. Then we try to regroup by "taking out" what is common.

We now rewrite the 2nd factor as such:

(x^3+1)

= x^3 + x^2 - x^2 + x - x + 1

= x^3 + x^2 - x^2 - x + x + 1

= x^2 (x+1) - x(x+1) + (x+1)

= (x^2 - x + 1) (x+1)

Combining with the first factor, we have:

**(x^3+5) (x^2 - x + 1) (x+1)**

To factorise x^6+6x^3+5.

We know that x^6 = ( x^3)^2.

Therefore we put x^3 = t. Then x^6 = t^2. Then the given exptression becomes aquadratic in t.

t^2 +6t+5.

t^2 +6t+5 = t^2+5t+t +5.

t^2 +6t+5 = t(t+5)+1(t+5).

t^2 +6t+5 = (t+5)(t+1).

Therefore x^6+6x^3 +5 = (x^3+5)(x^3+1).