Find the expansion of (3 x^{2}-2 a x+3 a^{2}) | vedclass

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(A) Using the Binomial Theorem,the given expression $(3 x^{2}-2 a x+3 a^{2})^{3}$ can be expanded as follows:$[(3 x^{2}-2 a x)+3 a^{2}]^{3}$$= {^3}C_0

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

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(A) Using the Binomial Theorem,the given expression $(3 x^{2}-2 a x+3 a^{2})^{3}$ can be expanded as follows:$[(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(9 x^{4}-12 a x^{3}+4 a^{2} x^{2})(3 a^{2}) + 3(3 x^{2}-2 a x)(9 a^{4}) + 27 a^{6}$$= (3 x^{2}-2 a x)^{3} + 81 a^{2} x^{4}-108 a^{3} x^{3}+36 a^{4} x^{2} + 81 a^{4} x^{2}-54 a^{5} x + 27 a^{6}$$= (3 x^{2}-2 a x)^{3} + 81 a^{2} x^{4}-108 a^{3} x^{3}+117 a^{4} x^{2}-54 a^{5} x + 27 a^{6} \quad \dots(1)$Now,expanding $(3 x^{2}-2 a x)^{3}$ using the Binomial Theorem:$(3 x^{2}-2 a x)^{3} = (3 x^{2})^{3} - 3(3 x^{2})^{2}(2 a x) + 3(3 x^{2})(2 a x)^{2} - (2 a x)^{3}$$= 27 x^{6} - 3(9 x^{4})(2 a x) + 3(3 x^{2})(4 a^{2} x^{2}) - 8 a^{3} x^{3}$$= 27 x^{6} - 54 a x^{5} + 36 a^{2} x^{4} - 8 a^{3} x^{3} \quad \dots(2)$Substituting $(2)$ into $(1)$:$= (27 x^{6} - 54 a x^{5} + 36 a^{2} x^{4} - 8 a^{3} x^{3}) + 81 a^{2} x^{4} - 108 a^{3} x^{3} + 117 a^{4} x^{2} - 54 a^{5} x + 27 a^{6}$$= 27 x^{6} - 54 a x^{5} + (36+81) a^{2} x^{4} - (8+108) a^{3} x^{3} + 117 a^{4} x^{2} - 54 a^{5} x + 27 a^{6}$$= 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}$

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

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