If a + b + c = 11 and ab + bc + ca = 38, then $$a^3 + b^3 + c^3 - 3abc$$ is equal to:

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

by using given data we can write has

$$11^2$$= $$a^2+b^2+c^2$$ + $$2 × 38$$

$$121$$= $$a^2+b^2+c^2$$ + $$76$$

$$121-76$$ = $$a^2+b^2+c^2$$

$$a^2+b^2+c^2$$= 45

$$a^3+b^3+c^3-3abc$$ =$$ (a+b+c)$$ $$(a^2+b^2+c^2-ab-bc-ca)$$

where $$(a+b+c)$$ = 11

$$a^2+b^2+c^2 $$= 45

and $$(ab+bc+ca)$$ = 38

Hence by applying to the formula

= $$11×(45-(38)$$

= $$11×(45-38)$$

= $$(11×7)$$

= 77