If A = a – b, B = b – c, C = c – a, then A3 + B3 + C3 – 3ABC is
Answers
Answer:
By assumption a^3+b^3+c^3=3abc so the left hand side is 0. Therefore (a+b+c)(a^2+b^2+c^2-ab-ac-bc) = 0
Answer:
This is an easy proof!
- A = a – b
- B = b – c
- C = c – a
We need to find:
A³+B³+C³-3ABC.
PROOF:
Substitute A = a – b, B = b – c, C = c – a in A³+B³+C³-3ABC.
(a−b)³+(b−c)³+(c−a)³−3(a−b)(b−c)(c−a)
Simplify the cubes.
=a³−3a²b+3ab²−b³+b³−3b²c+3bc²−c³−a³+3a²c−3ac²+c³+3a²b−3a²c−3ab²+3ac²+3b²c−3bc²
Combine Like Terms and rearrange:
=(a³−a³)+(−3a²b+3a²b)+(3a²c−3a²c)+(3ab²−3ab²)+(−3ac²+3ac²)+(−b3+b3)+(−3b²c+3b²c)+(3bc²−3bc²)+(−c³+c³)
You can clearly see that all the terms in the brackets will get cancelled. Thus:
(a³−a³)+(−3a²b+3a²b)+(3a²c−3a²c)+(3ab²−3ab²)+(−3ac²+3ac²)+(−b3+b3)+(−3b²c+3b²c)+(3bc²−3bc²)+(−c³+c³) = 0.
∴A³+B³+C³-3ABC=0.
ANOTHER METHOD IS:
Add A+B+C.
a – b + b – c + c – a = 0.
Thus, A+B+C=0. If A+B+C=0, then A³+B³+C³-3ABC=0 [This question is given on NCERT Class 9 textbook pg. 49 question 13. The answer to the 13th question is given in the file attached. Using the proof in the file attached, we can conclude that A³+B³+C³-3ABC=0.]
HOPE THIS HELPS :D
