Question

If a + b + c = 27, then what is the value of (a – 7)3 + (b – 9)3 + (c – 11)3 – 3(a – 7)(b – 9)(c – 11)?

Answer 1

a+b+c=27
(a-7)+(b-9)+(c-11)=27-7-9-11
(a-7)+(b-9)+(c-11)=0
So (a – 7)3 + (b – 9)3 + (c – 11)3 – 3(a – 7)(b – 9)(c – 11) = {(a-7)(b-9)(c-11)}^2{(a-7)^2+(b-9)^2+(c-11)^2}
0

Answer 2

Answer:

0

Step-by-step explanation:

a + b + c = 27 (Given)

We need to prove that - if a + b + c = 0, then a³ + b³+ c³ - 3abc = 0.

= (a + b + c)³ = a³ + b³ + c³ + 3a²b + 3ab² + 3b²c + 3bc² + 3c²a + 3ca² +6abc

= a³ + b³ + c³ + 3(ab + bc + ac)(a + b + c) - 3abc

Since a + b + c = 0, so a³ +b³ +c³ - 3abc = 0 (Q.E.D)

Now, a + b + c = 27 = (a - 7) + (b - 9) +(c - 11) = 0.

Thus, (a - 7)³ + (b - 9)³ + (c - 11)³ - 3(a - 7)(b - 9)(c - 11) =0.

Therefore the value of (a – 7)3 + (b – 9)3 + (c – 11)3 – 3(a – 7)(b – 9)(c – 11) is 0.