1. **State the problem:** Factor the polynomial expression $$6a^6 - 7a^3 b^3 - 8b^6$$. 2. **Identify the structure:** This is a trinomial in terms of $a^3$ and $b^3$. Let $x = a^3$ and $y = b^3$, so the expression becomes $$6x^2 - 7xy - 8y^2$$. 3. **Use factoring of quadratic form:** We want to factor $$6x^2 - 7xy - 8y^2$$ as $$(mx + ny)(px + qy)$$ where $m, n, p, q$ are constants. 4. **Find factors of $6 imes (-8) = -48$ that sum to $-7$:** The pair is $8$ and $-6$ because $8 + (-6) = 2$ (not correct), try $-12$ and $4$ (sum $-8$), try $-8$ and $6$ (sum $-2$), try $-16$ and $3$ (sum $-13$), try $-24$ and $2$ (sum $-22$), try $-4$ and $12$ (sum $8$), try $-3$ and $16$ (sum $13$), try $-1$ and $48$ (sum $47$). None sum to $-7$. So try factoring by grouping. 5. **Factor by grouping:** Rewrite as $$6x^2 - 12xy + 5xy - 8y^2$$. 6. **Group terms:** $$(6x^2 - 12xy) + (5xy - 8y^2)$$. 7. **Factor each group:** $$6x(x - 2y) + y(5x - 8y)$$. 8. **No common binomial factor, so try another approach:** Try to factor as $$(2x + ky)(3x + ly)$$. 9. **Expand:** $$(2x)(3x) + (2x)(ly) + (ky)(3x) + (ky)(ly) = 6x^2 + 2lxy + 3kxy + kly^2$$. 10. **Match coefficients:** - Coefficient of $x^2$ is $6$ (matches) - Coefficient of $xy$ is $2l + 3k = -7$ - Coefficient of $y^2$ is $kl = -8$ 11. **Find integer pairs $(k,l)$ such that $kl = -8$:** Possible pairs: $(1,-8), (-1,8), (2,-4), (-2,4), (4,-2), (-4,2), (8,-1), (-8,1)$. 12. **Test pairs to satisfy $2l + 3k = -7$:** - For $(k,l) = (1,-8)$: $2(-8) + 3(1) = -16 + 3 = -13$ - For $(k,l) = (-1,8)$: $2(8) + 3(-1) = 16 - 3 = 13$ - For $(k,l) = (2,-4)$: $2(-4) + 3(2) = -8 + 6 = -2$ - For $(k,l) = (-2,4)$: $2(4) + 3(-2) = 8 - 6 = 2$ - For $(k,l) = (4,-2)$: $2(-2) + 3(4) = -4 + 12 = 8$ - For $(k,l) = (-4,2)$: $2(2) + 3(-4) = 4 - 12 = -8$ - For $(k,l) = (8,-1)$: $2(-1) + 3(8) = -2 + 24 = 22$ - For $(k,l) = (-8,1)$: $2(1) + 3(-8) = 2 - 24 = -22$ No pair sums to $-7$. So try swapping $k$ and $l$. 13. **Try $2k + 3l = -7$ with $kl = -8$:** - For $(k,l) = (1,-8)$: $2(1) + 3(-8) = 2 - 24 = -22$ - For $(k,l) = (-1,8)$: $2(-1) + 3(8) = -2 + 24 = 22$ - For $(k,l) = (2,-4)$: $2(2) + 3(-4) = 4 - 12 = -8$ - For $(k,l) = (-2,4)$: $2(-2) + 3(4) = -4 + 12 = 8$ - For $(k,l) = (4,-2)$: $2(4) + 3(-2) = 8 - 6 = 2$ - For $(k,l) = (-4,2)$: $2(-4) + 3(2) = -8 + 6 = -2$ - For $(k,l) = (8,-1)$: $2(8) + 3(-1) = 16 - 3 = 13$ - For $(k,l) = (-8,1)$: $2(-8) + 3(1) = -16 + 3 = -13$ No pair sums to $-7$. 14. **Since no integer factorization works, try factoring the original expression by extracting common factors:** 15. **Rewrite original expression:** $$6a^6 - 7a^3 b^3 - 8b^6$$. 16. **Try to factor as a product of two cubics:** 17. **Try $(2a^3 + mb^3)(3a^3 + nb^3)$:** 18. **Expand:** $$6a^6 + 2n a^3 b^3 + 3m a^3 b^3 + mn b^6$$. 19. **Match coefficients:** - $6a^6$ matches - $2n + 3m = -7$ - $mn = -8$ 20. **Find integer pairs $(m,n)$ with $mn = -8$ and $2n + 3m = -7$:** Try $m=1, n=-8$: $2(-8) + 3(1) = -16 + 3 = -13$ Try $m=-1, n=8$: $2(8) + 3(-1) = 16 - 3 = 13$ Try $m=2, n=-4$: $2(-4) + 3(2) = -8 + 6 = -2$ Try $m=-2, n=4$: $2(4) + 3(-2) = 8 - 6 = 2$ Try $m=4, n=-2$: $2(-2) + 3(4) = -4 + 12 = 8$ Try $m=-4, n=2$: $2(2) + 3(-4) = 4 - 12 = -8$ Try $m=8, n=-1$: $2(-1) + 3(8) = -2 + 24 = 22$ Try $m=-8, n=1$: $2(1) + 3(-8) = 2 - 24 = -22$ No integer solution. 21. **Try rational factors:** Try $m = -rac{1}{2}, n = 4$: $2(4) + 3(-rac{1}{2}) = 8 - rac{3}{2} = rac{16}{2} - rac{3}{2} = rac{13}{2} eq -7$ Try $m = -rac{4}{3}, n = 3$: $2(3) + 3(-rac{4}{3}) = 6 - 4 = 2 eq -7$ 22. **Since no simple factorization exists, the expression is prime over integers.** **Final answer:** The polynomial $$6a^6 - 7a^3 b^3 - 8b^6$$ cannot be factored further using integer coefficients.