Test Index

NCERT Class XI Mathematics - Binomial Theorem - Solutions

© examsnet.com
Question : 36 of 36
Marks: +1, -0
Find the expansion of (3x2−2ax+3a2)3(3x^2 - 2ax + 3a^2)^3 using binomial theorem.
Solution:  
Let 3x23x^2 - 2ax = y
Then [3x2−2ax+3a2]3\left[3x^2-2ax+3a^2\right]^3 = [y+3a2]3\left[y+3a^2\right]^3
= 3C0y3+ 3C1y2(3a2)+ 3C2y(3a2)2+ 3C3(3a2)3= \, {}^3C_0 y^3 + \, {}^3C_1 y^2(3a^2) + \, {}^3C_2 y (3a^2)^2 + \, {}^3C_3 (3a^2)^3
=(3x2−ax)3+3(3x2−2ax)2(3a2)+3(3x2−2ax)(9a4)+(27a6)=(3x^2-ax)^3 + 3(3x^2-2ax)^2(3a^2) + 3(3x^2-2ax)(9a^4)+(27a^6)
=[ 3C0(3x2)3+ 3C1(3x2)2(−2ax)+ 3C2(3x2)(−2ax)2+ 3C3(−2ax)3]=\left[\, {}^3C_0(3x^2)^3 + \, {}^3C_1(3x^2)^2(-2ax) + \, {}^3C_2(3x^2)(-2ax)^2 + \, {}^3C_3(-2ax)^3\right]
+
3[ 2C0(3x2)2+ 2C1(3x2)(−2ax)+ 2C2(−2ax)2](3a2)+3(27x2a4−18a5x)+27a63\left[\, {}^2C_0(3x^2)^2 + \, {}^2C_1(3x^2)(-2ax) + \, {}^2C_2(-2ax)^2\right] (3a^2) + 3(27x^2 a^4-18a^5 x) + 27a^6
= 27x6−54ax5+36a2x427x^6 - 54ax^5 + 36a^2x^4 - 8a3x3+81a2x48a^3 x^3 +81a^2x^4 - 108a3x3+36a4x2+81a4x2108a^3 x^3 + 36a^4x^2+81a^4x^2 - 54a5x+27a654a^5x+27a^6
= 27x6−64ax5+117a2x427x^6-64ax^5+117a^2x^4 - 116a3x3116a^3x^3 + 117a4x2−54a5x+27a6117a^4x^2-54a^5x+27a^6
© examsnet.com
Go to Question:
Prev Question