If a + b + c = 2, $$a^2 + b^2 + c^2 = 26$$, then the value of $$a^3 + b^3 + c^3 - 3 abc$$ is:

By using this formula we can solve

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

by using given data we can write has

$$2^2$$=$$26$$+$$2(ab+bc+ca)$$

$$4-26÷2$$= $$(ab+bc+ca)$$

$$-22÷2$$ =  $$(ab+bc+ca)$$

$$(ab+bc+ca)$$ = -11

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

where $$(a+b+c)$$ = 2

$$a^2+b^2+c^2 $$= 26

and $$(ab+bc+ca)$$ = -11

Hence by applying to the formula

= $$2×(26-(-11)$$

= $$2×(26+11)$$

= $$2×(37)$$

= 74