Answer:
[tex]27c^3-27c^2d+9cd^2-d^3[/tex]
Step-by-step explanation:
Binomial Theorem
[tex](a+b)^n=a^n+\dfrac{n!}{1!(n-1)!}a^{n-1}b+\dfrac{n!}{2!(n-2)!}a^{n-2}b^2+...+\dfrac{n!}{r!(n-r)!}a^{n-r}b^r+...+b^n[/tex]
[tex]\textsf{If }(a+b)^n=(3c-d)^3 \: \textsf{ then}:[/tex]
Substitute the given values into the formula:
[tex]\begin{aligned}(3c-d)^3 & =(3c)^3+\dfrac{3!}{1!(3-1)!}(3c)^{3-1}(-d)+\dfrac{3!}{2!(3-2)!}(3c)^{3-2}(-d)^2+(-d)^3\\\\& =(3c)^3+3(3c)^2(-d)+3(3c)^1(-d)^2+(-d)^3\\\\& =27c^3-27c^2d+9cd^2-d^3\end{aligned}[/tex]
