If the sum of two numbers is 11 and the sum of their squares is 65, then the sum of their cubes will be:

Let the two numbers are $$a$$ and $$b$$

Given, Sum of their squares = 65

$$=$$>  $$a^2+b^2=65$$ ...................(1)

Sum of two numbers = 11

$$=$$>  $$a+b=11$$

$$=$$>  $$\left(a+b\right)^2=11^2$$

$$=$$>  $$a^2+b^2+2ab=121$$

$$=$$>  $$65+2ab=121$$

$$=$$>  $$2ab=56$$

$$=$$>  $$ab=28$$ ............................(2)

$$\therefore\ $$Sum of their cubes = $$a^3+b^3$$

$$=\left(a+b\right)\left(a^2+b^2-ab\right)$$

$$=\left(11\right)\left(65-28\right)$$

$$=\left(11\right)\left(37\right)$$

$$=407$$

Hence, the correct answer is Option A