If $a + b + c = 9$ then find the value of
$(3 - a)^{3} + (3 - b)^{3} + (3 - c)^{3} - 3 (3 - a) (3 - b) (3 - c)$

Open in App

Solution

Given that,
$a + b + c = 9$
$∴ {(3 - a)}^{3} + {(3 - b)}^{3} + {(3 - c)}^{3} - 3 (3 - a) (3 - b) (3 - c)$
$= (3 - a + 3 - b + 3 - c) ∣ ∣ {(3 - a)}^{2} + {(3 - b)}^{2} + {(3 - c)}^{2} - (3 - a) (3 - b) - (3 - b) (3 - c) - (3 - c) (3 - a) ∣ ∣$
$| 9 - (a + b + c) | ∣ ∣ {(3 - a)}^{2} + {(3 - b)}^{2} + {(3 - c)}^{2} - (3 - a) (3 - b) - (3 - b) (3 - c) - (3 - c) (3 - a) ∣ ∣$
$(9 - 9) ∣ ∣ {(3 - a)}^{2} + {(3 - b)}^{2} + {(3 - c)}^{2} - (3 - a) (3 - b) - (3 - b) (3 - c) - (3 - c) (3 - a) ∣ ∣$
$0 ∣ ∣ {(3 - a)}^{2} + {(3 - b)}^{2} + {(3 - c)}^{2} - (3 - a) (3 - b) - (3 - b) (3 - c) - (3 - c) (3 - a) ∣ ∣$
$= 0$

Suggest Corrections

Similar questions

> 0

\frac{a^{3}}{b^{3}} + \frac{b^{3}}{c^{3}} + \frac{c^{3}}{a^{3}} \geq 3

If a + b + c = 9, and $a^{2} + b^{2} + c^{2} = 35$ , find the value of $a^{3} + b^{3} + c^{3} - 3 a b c .$

a + b + c = 0

(a + b - c)^{3} + (b + c - a)^{3} + (c + a - b)^{3}

If $a + b + c = 0$ , then the value of $(a + b - c)^{3} + (b + c - a)^{3} + (c + a - b)^{3}$ is:

a^{3} (b - c)^{3} + b^{3} (c - a)^{3} + c^{3} (a - b)^{3}

Join BYJU'S Learning Program