First separate out the common scalar factor 6 To Find.

Then notice that both x³ And 8=2³ are perfect cubes, so work well with the sum of cubes identity.

A³+B³=(A+B)(A²-AB+B²)

With A=x And B=2 We Find.

Putting it together we get.

This has no simpler factors with Real coefficients, as you can check by looking at the discriminant

Δ Of (x²-2x+4)

Δ=b²-4ac=(-2)²-(4×+×4)

Since Δ<0 this quadratic has no Real zeros and no linear factors with Real coefficients.

D.x²-2x+4

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