If a + b + c = 9 and ab + bc + ca = 18, then what is the value of $$a^3 + b^3 + c^3 - 3abc$$ ?

Given : $$ab+bc+ca=18$$ ---------------(i)

and $$a+b+c=9$$ --------------(ii)

Squaring both sides, we get :

=> $$(a+b+c)^2=(9)^2$$

=> $$(a^2+b^2+c^2)+2(ab+bc+ca)=81$$

=> $$a^2+b^2+c^2+2(18)=81$$

=> $$a^2+b^2+c^2=81-36=45$$ ------------(iii)

To find : $$a^3 + b^3 + c^3 - 3abc$$

= $$(a+b+c)(a^2+b^2+c^2-ab-bc-ca)$$

Substituting values from equations (i),(ii) and (iii),

= $$(9)(45-18)$$

= $$9\times27=243$$

=> Ans - (B)