# Find the expansion of using binomial theorem.

Question:

Find the expansion of $\left(3 \mathrm{x}^{2}-2 \mathrm{ax}+3 \mathrm{a}^{2}\right)^{3}$ using binomial theorem.

Solution:

Using Binomial Theorem, the given expression $\left(3 x^{2}-2 a x+3 a^{2}\right)^{3}$ can be expanded as

$\left[\left(3 x^{2}-2 a x\right)+3 a^{2}\right]^{3}$

$={ }^{3} C_{0}\left(3 x^{2}-2 a x\right)^{3}+{ }^{3} C_{1}\left(3 x^{2}-2 a x\right)^{2}\left(3 a^{2}\right)+{ }^{3} C_{2}\left(3 x^{2}-2 a x\right)\left(3 a^{2}\right)^{2}+{ }^{3} C_{3}\left(3 a^{2}\right)^{3}$

$=\left(3 x^{2}-2 a x\right)^{3}+3\left(9 x^{4}-12 a x^{3}+4 a^{2} x^{2}\right)\left(3 a^{2}\right)+3\left(3 x^{2}-2 a x\right)\left(9 a^{4}\right)+27 a^{6}$

$=\left(3 x^{2}-2 a x\right)^{3}+81 a^{2} x^{4}-108 a^{3} x^{3}+36 a^{4} x^{2}+81 a^{4} x^{2}-54 a^{5} x+27 a^{6}$

$=\left(3 x^{2}-2 a x\right)^{3}+81 a^{2} x^{4}-108 a^{3} x^{3}+117 a^{4} x^{2}-54 a^{5} x+27 a^{6}$ (1)

Again by using Binomial Theorem, we obtain

$\left(3 x^{2}-2 a x\right)^{3}$

$={ }^{3} C_{0}\left(3 x^{2}\right)^{3}-{ }^{3} C_{1}\left(3 x^{2}\right)^{2}(2 a x)+{ }^{3} C_{2}\left(3 x^{2}\right)(2 a x)^{2}-{ }^{3} C_{1}(2 a x)^{3}$

$=27 x^{6}-3\left(9 x^{4}\right)(2 a x)+3\left(3 x^{2}\right)\left(4 a^{2} x^{2}\right)-8 a^{3} x^{3}$

$=27 x^{6}-54 a x^{5}+36 a^{2} x^{4}-8 a^{3} x^{3}$ $\ldots(2)$

From (1) and (2), we obtain

$\left(3 x^{2}-2 a x+3 a^{2}\right)^{3}$

$=27 \mathrm{x}^{6}-54 \mathrm{ax}^{5}+36 \mathrm{a}^{2} \mathrm{x}^{4}-8 \mathrm{a}^{3} \mathrm{x}^{3}+81 \mathrm{a}^{2} \mathrm{x}^{4}-108 \mathrm{a}^{3} \mathrm{x}^{3}+117 \mathrm{a}^{4} \mathrm{x}^{2}-54 \mathrm{a}^{5} \mathrm{x}+27 \mathrm{a}^{6}$

$=27 x^{6}-54 a x^{5}+117 a^{2} x^{4}-116 a^{3} x^{3}+117 a^{4} x^{2}-54 a^{5} x+27 a^{6}$