Question

Find the expansion of \((3x^2 – 2ax + 3a^2)^3\) using binomial theorem.

Answer 1

Using Binomial Theorem, the given expression \((3x^2-2ax +3a^2)^3\) can be expanded as
\([(3x^2-2ax)+3a^2]^3\)
= \(^3C_0 (3x^2 -2ax)^3 + \space^3C_1(3x^2 - 2ax)^2 (3a^2) +\space ^3C_2 (3x^2 - 2ax) (3a^2)^2 + \space^3C_3 (3a^2)^2\)
\(=(3x^2-2ax)^3+3(9x^4-12ax^3 +4a^2x^2) (3a^2)+3(3x^2-2ax) (9a^4)+27a^6\)
\(=(3x^2-2ax)^3 +81a^2x^4-108a^3x^3 +36a^4x^2+81a^4x^2-54a^5x+27a^6\)
\(=(3x^2-2ax)^3 +81a^2x^4-108a^3x^3+117a^4x^2-54a^5x+27a^6          ...(1)\)

Again by using Binomial Theorem, we obtain
\((3x^2-2ax)^3\)
\(=\space^ 3C_0 (3x^2)^3 - \space^3C_1 (3x^2)^2 (2ax) + \space^3C_2 (3x^2) (2ax)^2 - 3C^3 (2ax)^3 \)
\(=27x^6-3(9x^4) (2ax)+3(3x^2) (4a^2x^2)-8a^3x^3\)
\(=27x^6-54ax^5 +36a^2x^4+-8a^3x^3                    ...(2)\)

From (1) and (2), we obtain
\((3x^2-2ax +3a^2)^3\)
\(=27x^6-54ax^5 +36a^2x^4-8a^3x^3 +81a^2x^4-108a^3x^3 +117a^4x^2 - 54a^5x+27a^6\)
\(=27x^6-54ax^5+117a^2x^4+-116a^3x^3 +117a^4x^2 -54a^5 +27a^6\)