We have a + b + c = 9 ...(i)
⇒ (a + b + c)2 = 81 [On squaring both sides of (i)]
⇒ a2 + b2 + c2 + 2(ab + bc + ac) = 81
⇒ a2 + b2 + c2 + 2 × 26 = 81 [∵ab + bc + ac = 26]
⇒ a2 + b2 + c2 = (81 - 52)
⇒ a2 + b2 + 2 = 29.
Now, we have
a3 + b3 + c3 - 3abc = (a + b + c) (a2 + b2 + c2 - ab - bc - ac)
= (a + b + c) [(a2 + b2 + c2 ) - (ab + bc + ac)]
= 9 × [(29 - 26)]
= (9 × 3)
= 27
