Answer:
(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:
The expansion is given by the following formula: (a+b)n=∑k=0n(nk)an−kbk,
where (nk)=n!(n−k)!k! and n!=1⋅2⋅3...n.
We have that a=3c, b=d2, n=6.
Therefore, (3c+d2)6=∑k=06(6k)(3c)6−k(d2)k
Now, calculate the product for every value of k from 0 to 6.
k=0: (60)(3c)6−0(d2)0=6!(6−0)!0!(3c)6(d2)0=729c6
k=1: (61)(3c)6−1(d2)1=6!(6−1)!1!(3c)5(d2)1=1458c5d2
k=2: (62)(3c)6−2(d2)2=6!(6−2)!2!(3c)4(d2)2=1215c4d4
k=3: (63)(3c)6−3(d2)3=6!(6−3)!3!(3c)3(d2)3=540c3d6
k=4: (64)(3c)6−4(d2)4=6!(6−4)!4!(3c)2(d2)4=135c2d8
k=5: (65)(3c)6−5(d2)5=6!(6−5)!5!(3c)1(d2)5=18cd10
k=6: (66)(3c)6−6(d2)6=6!(6−6)!6!(3c)0(d2)6=d12
Finally, (3c+d2)6=∑k=06(6k)(3c)6−k(d2)k=(60)(3c)6−0(d2)0+(61)(3c)6−1(d2)1+(62)(3c)6−2(d2)2+(63)(3c)6−3(d2)3+(64)(3c)6−4(d2)4+(65)(3c)6−5(d2)5+(66)(3c)6−6(d2)6=729c6+1458c5d2+1215c4d4+540c3d6+135c2d8+18cd10+d12
Answer is above :)
