If $$a + b + c = 9, ab + bc + ca = 26, a^3 + b^3= 91, b^3 + c^3 = 72$$ and $$c^3 + a^3 = 35$$, then what is the value of $$abc$$?

We know that :

$$a^3+b^3+c^3=\left(a+b+c\right)\left(a^2+b^2+c^2-ab-ac-bc\right)+3abc\ .$$

Here, $$a + b + c = 9, ab + bc + ca = 26, a^3 + b^3= 91, b^3 + c^3 = 72$$ and $$c^3 + a^3 = 35$$

So, $$2\left(a^3+b^3+c^3\right)=72+91+35\ =198.$$

So, $$a^3+b^3+c^3=99\ .$$

And , $$a^2+b^2+c^2=9^2-2\times26=29\ .$$

So, $$a^3+b^3+c^3=\left(a+b+c\right)\left(a^2+b^2+c^2-ab-ac-bc\right)+3abc\ $$

or, $$99=\left(9\right)\left(29-26\right)+3abc\ .$$

or, $$abc=24\ .$$

B is correct choice.