Answer: a³+b³+c³-3abc = 27

Step-by-step explanation: Let us list what has been given to us:

a+b+c=9

ab+bc+ca=26

We shall use:

(a+b+c)² = a²+b²+2(ab+bc+ca)

Now let us substitute the values

(9)² = a²+b²+c²+2(26)

Therefore, 81-52 = a²+b²+c²

Therefore, a²+b²+c² = 29

Now let us use:

a³+b³+c³-3abc=(a+b+c)(a²+b²+c²-ab-bc-ca)

Therefore, a³+b³+c³-3abc=(a+b+c)[(a²+b²+c²)-(ab+bc+ca)]

Now let us substitute a+b+c=9, ab+bc+ca=26 and a²+b²+c²=29

Therefore,

a³+b³+c³-3abc = 9[(29-26)]

=9×3

=27

Hope this helps!