Expand #(2x+3)^3# using binomial expansion?

1 Answer
Shwetank Mauria
Apr 28, 2017

#(2x+3)^3=8x^3+12x^2+54x+27#

Explanation:

Binomial expansion of #(a+b)^n=C_0^na^n+C_1^na^(n-1)b+C_2^na^(n-2)b^2+C_3^na^(n-3)b^3+..++C_n^n)b^n#

where #C_r^n=(n(n-1)(n-2).....(n-r+1))/(1.2.3.....r)# and #C_0^n=1#

Hence #(2x+3)^3#

= #C_0^3(2x)^3+C_1^3(2x)^(3-1)3+C_2^3(2x)^(3-2)3^2+C_3^3(2x)^(3-3)3^3#

= #(2x)^3+3/1(2x)^2xx3+(3xx2)/(1xx2)(2x)9+(3xx2xx1)/(1xx2xx3)(2x)^0xx27#

= #8x^3+12x^2+54x+27#

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