Factor by grouping \( 12 x^{6}-8 x^{3}-15 x^{3}+12 \)

To factor the expression \( 12x^6 - 8x^3 - 15x^3 + 12 \) by grouping, we can first rearrange the terms: \[ 12x^6 + (-8x^3 - 15x^3) + 12 \] This simplifies to: \[ 12x^6 - 23x^3 + 12 \] Next, we can group the terms into two pairs: \[ (12x^6 + 12) + (-23x^3) \] Now, we can factor out the common factors from each group: 1. From the first group \( 12x^6 + 12 \), we can factor out \( 12 \): \[ 12(x^6 + 1) \] 2. The second group \( -23x^3 \) does not have a common factor with the first group, so we leave it as is. Now we have: \[ 12(x^6 + 1) - 23x^3 \] Next, we can try to factor \( x^6 + 1 \) further. Notice that \( x^6 + 1 \) can be factored using the sum of cubes: \[ x^6 + 1 = (x^2)^3 + 1^3 = (x^2 + 1)((x^2)^2 - (x^2)(1) + 1^2) = (x^2 + 1)(x^4 - x^2 + 1) \] Now substituting this back into our expression gives: \[ 12((x^2 + 1)(x^4 - x^2 + 1)) - 23x^3 \] At this point, we can write the entire expression as: \[ 12(x^2 + 1)(x^4 - x^2 + 1) - 23x^3 \] However, we cannot factor this expression further in a simple way. Thus, the factored form of the original expression is: \[ 12(x^2 + 1)(x^4 - x^2 + 1) - 23x^3 \] This is the factored form by grouping.