Answer:
Option (a) is correct.
\((3c+d^2)^6=729c^6+1458c^5d^2+1215c^4d^4+540c^3d^6+135c^2d^8+18cd^{10}+d^{12}\)
Step-by-step explanation:
Given : \(\left(3c+d^2\right)^6\)
We have to expand the given expression and choose the correct from the given options.
Consider the given expression \(\left(3c+d^2\right)^6\)
Using binomial theorem ,
\(\left(a+b\right)^n=\sum _{i=0}^n\binom{n}{i}a^{\left(n-i\right)}b^i\)
We have \(a=3c,\:\:b=d^2\)
\(=\sum _{i=0}^6\binom{6}{i}\left(3c\right)^{\left(6-i\right)}\left(d^2\right)^i\)
also, \(\binom{n}{i}=\frac{n!}{i!\left(n-i\right)!}\)
for i = 0 , we have,
\(\frac{6!}{0!\left(6-0\right)!}\left(3c\right)^6d^2^0=729c^6\)
for i = 1 , we have,
\(\frac{6!}{1!\left(6-1\right)!}\left(3c\right)^5d^2^1=1458c^5d^2\)
for i = 2 , we have,
\(\frac{6!}{2!\left(6-2\right)!}\left(3c\right)^4d^2^2=1215c^4d^4\)
for i = 3 , we have,
\(\frac{6!}{3!\left(6-3\right)!}\left(3c\right)^3d^2^3=540c^3d^6\)
for i = 4 , we have,
\(\frac{6!}{4!\left(6-4\right)!}\left(3c\right)^2d^2^4=135c^2d^8\)
for i = 5 , we have,
\(\frac{6!}{4!\left(6-4\right)!}\left(3c\right)^2d^2^4=18cd^{10}\)
for i = 6 , we have,
\(\frac{6!}{6!\left(6-6\right)!}\left(3c\right)^0d^2^6=d^{12}\)
Thus, adding all term together, we have,
\((3c+d^2)^6=729c^6+1458c^5d^2+1215c^4d^4+540c^3d^6+135c^2d^8+18cd^{10}+d^{12}\)
Thus, Option (a) is correct.
