The key idea is to treat the trinomial as a binomial by grouping two terms together, then expand using the Binomial Theorem twice. The final expansion is 27x6−54ax5+117a2x4−116a3x3+117a4x2−54a5x+27a6.
Why This Approach Works
The Binomial Theorem directly handles expressions of the form (A+B)n, but here we have three terms inside the cube. The natural trick is to group any two terms together as a single "binomial" — say, let A=3x2−2ax and B=3a2. Then (A+B)3 expands via the binomial theorem, and each term in that expansion itself contains a binomial (3x2−2ax) raised to a power, which we expand again. This two-stage expansion is systematic and avoids errors.
A common mistake is to try to apply the binomial theorem directly to three terms without grouping — that leads to missing cross-terms. Always group first.
Step-by-Step Expansion
1. Group the trinomial as a binomial.
Write (3x2−2ax+3a2)3=[(3x2−2ax)+3a2]3.
2. Apply the Binomial Theorem to the grouped form.
Recall: (A+B)3=A3+3A2B+3AB2+B3.
Here A=3x2−2ax and B=3a2. So:
(3x2−2ax+3a2)3=(3x2−2ax)3+3(3x2−2ax)2(3a2)+3(3x2−2ax)(3a2)2+(3a2)3
3. Simplify the constant factors.
Notice each term has powers of a and x that we'll track carefully. Compute the coefficients:
- 3(3a2)=9a2 for the second term.
- 3(3a2)2=3⋅9a4=27a4 for the third term.
- (3a2)3=27a6 for the fourth term.
So the expansion becomes:
(3x2−2ax)3+9a2(3x2−2ax)2+27a4(3x2−2ax)+27a6
4. Expand (3x2−2ax)3.
Treat C=3x2 and D=−2ax. Then (C+D)3=C3+3C2D+3CD2+D3.
- C3=(3x2)3=27x6
- 3C2D=3(3x2)2(−2ax)=3⋅9x4⋅(−2ax)=−54ax5
- 3CD2=3(3x2)(−2ax)2=3⋅3x2⋅4a2x2=36a2x4
- D3=(−2ax)3=−8a3x3
Thus:
(3x2−2ax)3=27x6−54ax5+36a2x4−8a3x3
5. Expand (3x2−2ax)2.
This is simpler: (3x2)2+2(3x2)(−2ax)+(−2ax)2=9x4−12ax3+4a2x2.
6. Multiply by 9a2 (from step 3).
9a2(3x2−2ax)2=9a2(9x4−12ax3+4a2x2)=81a2x4−108a3x3+36a4x2
7. Expand 27a4(3x2−2ax).
27a4(3x2−2ax)=81a4x2−54a5x
8. Add the constant term 27a6.
Now combine all four pieces: