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

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(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

Vedclass · Accepted Answer

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

Vedclass · Answer

(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.

Similar Questions

The greatest integer less than or equal to $(\sqrt{3}+2)^5$ is

If $x$ is so small that all terms containing $x^2$ and higher powers of $x$ can be neglected, then the approximate value of $\frac{(1+\frac{2x}{3})^{-4}(4+5x)^{1/2}}{(9+x)^{3/2}}$ when $x=\frac{6}{371}$, is

The last two digits of the number $3^{400}$ are:

The coefficient of $x^{48}$ in $(1+x)+2(1+x)^2+3(1+x)^3+ . . . +100(1+x)^{100}$ is equal to:

If $(1+x+x^2+x^3)^5 = \sum_{k=0}^{15} a_k x^k$,then $\sum_{k=0}^7 a_{2k}$ is equal to

Vedclass Test Series

Exam Paper Generator

Online Exam Module