Question

3. (2x - 3)®6
expand
​

Answer 1

$\large\sf{\underline{To\: Expand-}}$

$\rm\implies \:{(2x - 3) }^{6}$

$\large\sf{\underline{Solution-}}$

We will solve this question by using binomial theorem.

According to binomial theorem,

$\underline{\large\boxed{\: \rm\:{\big(a + b\big) }^{n} ={^nC_0.a^n.b^0}+{^nC_1.a^{n-1}.b^1}+...+{^nC_na^{n-n}.b^n}}}$

or

$\underline{\boxed{\rm(a+b)^n=\sum\limits^n_{k=0}{^nC_k}\quad\,a^{n-k}\;b^k}}$

By applying this formula we get,

$\small\rm\to\big(2x+\{-3\}\big)^6={^6C_0}.(2x)^6.{(-3)}^0+{^6C_1}.{(2x)}^5(-3)^1+{^6C_2.(2x)^4.(-3)^2}+{^6C_3.(2x)^3.(-3)^3}+{^6C_4.(2x)^2.(-3)^4}+{^6C_5.(2x)^1.(-3)^5}+{^6C_6.(2x)^0.(-3)^6}$

Now we have formula $\small\rm{^nC_r=\dfrac{n!}{(n-r)!(r)!}}$

Here n! means n×(n-1) × (n-2) × . . . × 3 × 2 × 1

Applying this we get,

$\rm\small \to(1) 64x^6+(6)(32x^5)(-3)+(15)(16x^4)(9)+(20)(8x^3)(-27)+(15)4x^2(81)+6(2x)(-3)^5+(-3)^6$

By simplifying it we get,

${\to \rm\small 64x^6-576x^5+2160x^4-4320x^3+4860x^2-2916x+729}$

So the required answer is,

$\green{ \boxed{ \boxed{\to \rm 64x^6-576x^5+2160x^4-4320x^3+4860x^2-2916x+729}}}$

Kindly see this answer from website for better understanding.

Thank you!