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Binomial Expression

Find the expa...

Question

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

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Solution

We have $(3 x^{2} - 2 a x + 3 a^{2})^{3}$

$= [(3 x^{2} - 2 a x) + 3 a^{2}]^{3}$

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

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

$= (27 x^{6} - 8 a^{3} x^{3} - 54 a x^{5} + 36 a^{2} x^{4}) + 9 a^{2} (9 x^{4} + 4 a^{2} x^{2} - 12 a x^{3}) + 27 a^{4} (3 x^{2} - 2 a x) + 27 a^{6}$

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

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

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