If (x + a)(x + b)(x + c) ≡ x³ – 10x² + 45x – 15 find a + b + c, and a² + b² + c².

Here, we have to use the identity

(x + a)(x + b)(x + c) = x³ + (a + b + c)x² + (ab + bc + ca)x + abc

Comparing this with

x³ – 10x² + 45x – 15 = x³ + (a + b + c)x² + (ab + bc + ca)x + abc

⇒ – 10 = a + b + c

Now,

And (ab + bc + ca) = 45

Now (a + b + c)² = a² + b² + c² + 2(ab + bc + ca)

⇒ (– 10)² – 2 × 45 = a² + b² + c²

⇒ 100 – 90 = a² + b² + c²

⇒ a² + b² + c² = 10