If a + b + c = 11 and ab + bc + ca = 17, then what is the value of $$a^3 + b^3 + c^3 - 3abc$$ ?

Given : $$ab+bc+ca=17$$ -------------(i)

and $$a+b+c=11$$ -----------------(ii)

Squaring both sides, we get :

=> $$(a+b+c)^2=(11)^2$$

=> $$(a^2+b^2+c^2)+2(ab+bc+ca)=121$$

=> $$(a^2+b^2+c^2)+2(17)=121$$

=> $$a^2+b^2+c^2=121-34=87$$ --------------(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), we get :

= $$(11)(87-17)$$

= $$11\times70=770$$

=> Ans - (D)