If a + b + c = 9 and ab + bc + ac = 26, find the value of a3 + b3 + c3

1.	If a + b + c = 9 and ab + bc + ac = 26, find the value of a3 + b3 + c3 - 3abc.
Answer» 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

If a + b + c = 9 and ab + bc + ac = 26, find the value of a3 + b3 + c3 - 3abc.

Answer»

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