Question 17 of 20advanced🔧 ApplyLong Answer5 marks

Find the expansion of (3x2−2ax+3a2)3(3x^2 - 2ax + 3a^2)^3 using binomial theorem.

Correct Answer

27x6−54ax5+117a2x4−116a3x3+117a4x2−54a5x+27a627x^6 - 54ax^5 + 117a^2x^4 - 116a^3x^3 + 117a^4x^2 - 54a^5x + 27a^6

Exercise: Miscellaneous Exercise on Chapter 7 | Q: 6 | (Chapter: 8)
For More Understanding

Explanation

To expand a trinomial using the binomial theorem, we group two terms together to form a binomial. The question tests understanding of binomial expansion and algebraic simplification.

Students must correctly apply the formula

(a+b)3=a3+3a2b+3ab2+b3(a+b)^3 = a^3 + 3a^2b + 3ab^2 + b^3

twice — first for the main expansion and then for expanding the grouped term (−2ax+3a2)3(-2ax + 3a^2)^3. Careful collection of like terms is essential for the final answer.

Solution Steps

  1. Step 1: Group the expression as [3x2+(−2ax+3a2)]3[3x^2 + (-2ax + 3a^2)]^3. Let A=3x2A = 3x^2 and B=−2ax+3a2B = -2ax + 3a^2.

  2. Step 2: Apply binomial theorem:

    (A+B)3=A3+3A2B+3AB2+B3(A + B)^3 = A^3 + 3A^2B + 3AB^2 + B^3

  3. Step 3: Calculate A3A^3:

    A3=(3x2)3=27x6A^3 = (3x^2)^3 = 27x^6

  4. Step 4: Calculate 3A2B3A^2B:

    3A2B=3(3x2)2(−2ax+3a2)=27x4(−2ax+3a2)=−54ax5+81a2x43A^2B = 3(3x^2)^2(-2ax + 3a^2) = 27x^4(-2ax + 3a^2) = -54ax^5 + 81a^2x^4

  5. Step 5: Calculate B2B^2:

    B2=(−2ax+3a2)2=4a2x2−12a3x+9a4B^2 = (-2ax + 3a^2)^2 = 4a^2x^2 - 12a^3x + 9a^4

  6. Step 6: Calculate 3AB23AB^2:

    3AB2=3(3x2)(4a2x2−12a3x+9a4)=36a2x4−108a3x3+81a4x23AB^2 = 3(3x^2)(4a^2x^2 - 12a^3x + 9a^4) = 36a^2x^4 - 108a^3x^3 + 81a^4x^2

  7. Step 7: Calculate B3B^3:

    B3=(−2ax+3a2)3=−8a3x3+36a4x2−54a5x+27a6B^3 = (-2ax + 3a^2)^3 = -8a^3x^3 + 36a^4x^2 - 54a^5x + 27a^6

  8. Step 8: Add all terms and collect like terms:

    27x6−54ax5+117a2x4−116a3x3+117a4x2−54a5x+27a627x^6 - 54ax^5 + 117a^2x^4 - 116a^3x^3 + 117a^4x^2 - 54a^5x + 27a^6

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