Explanation
We know, a perfect cube has multiples of $$3$$ as powers of prime factors.
Here, number of $$3$$'s is $$3$$ and number of $$37$$'s is $$1$$.
Given number $$216$$ is the cube of $$6$$, i.e. $$6^3=216$$.
We know,
$$\displaystyle 1=1={ 1 }^{ 3 }$$,
$$\displaystyle 3+5=8={ 2 }^{ 3 }$$,
$$\displaystyle 7+9+11=27={ 3 }^{ 3 }$$,
$$\displaystyle 13+15+17+19=64={ 4 }^{ 3 }$$,
$$\displaystyle 21+23+25+27+29=125={ 5 }^{ 3 }$$,
$$\displaystyle 31+33+35+37+39+41=216={ 6 }^{ 3 }$$.
$$\displaystyle \therefore $$ The sum of six consecutive odd numbers are to form $$216$$ is $$\displaystyle 31+33+35+37+39+41=216$$.
Hence, option $$D$$ is correct.
Given number $$64$$ is the cube of $$4$$, i.e. $$4^3=64$$.
$$\displaystyle 13+15+17+19=64={ 4 }^{ 3 }$$.
$$\displaystyle \therefore $$ The sum of four consecutive odd numbers are to form $$64$$ is $$\displaystyle 13+15+17+19=64$$.
Hence, option $$B$$ is correct.
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