If the sum of three numbers is 18 and the sum of their squares is 36, then find the difference between the sum of their cubes and three times of their product.

Correct option is B

Given:

Sum of three numbers =18

Sum of their squares  = 36

Formula Used:

​$$a^{3}+b^{3}+c^{3}-3abc = (a+b+c)(a^{2}+b^{2}+c^{2}-ab-bc-ac)$$​​

​$$(a+b+c)^{2}= a^{2}+b^{2}+c^{2} +2(ab+bc+ac)$$​​

Solution:

Let the number be $$x,y,z$$​​

Then according to question:

​$$x+y+z =18$$​​

​$$x^{2}+y^{2}+z^{2} =36$$​​

​$$a^{3}+b^{3}+c^{3}-3abc = (a+b+c)(a^{2}+b^{2}+c^{2}-ab-bc-ac)$$​​

For using this formula,, we need to find value of (ab+bc+ca)

For that we use $$x+y+z =18$$​​

Squaring both sides

​$$(x+y+z)^{2}= 18^{2}$$​​

​$$x^{2}+y^{2}+z^{2} +2(xy+yz+xz) =324$$​​

​$$36 +2(xy+yz+xz) =324$$​    

​$$2(xy+yz+xz) =324-36$$​​

​$$(xy+yz+xz) = 144$$​​

​$$x^{3}+y^{3}+z^{3}-3xyz = (x+y+z)(x^{2}+y^{2}+z^{2}-xy-yz-xz)$$​​

$$x^{3}+y^{3}+z^{3}-3xyz = (18)(36-(xy+yz+xz))$$​

​$$x^{3}+y^{3}+z^{3}-3xyz = (18)(-108)$$​​

​$$x^{3}+y^{3}+z^{3}-3xyz = -1944$$​​