A+b+c=0 find a2/bc+b2/ac+c2/ab
Answers
Answer:
3
Step-by-step explanation:
Given;
a + b + c = 0
(a^2/bc) + (b^2)/ac + (c^2)/ab = (a^3 + b^3 + c^3)/abc
[We know that,
(a + b + c)^3 = a^3 + b^3 + c^3 + 3ab(a + b + c) + 3bc(a + b + c) + 3ac(a + b + c) - 3abc
So,
a^3 + b^3 + c^3 = (a + b + c)^3 - 3ab(a + b + c) - 3bc(a + b + c) - 3ac(a + b + c) + 3abc ]
(a^2/bc) + (b^2)/ac + (c^2)/ab = (a^3 + b^3 + c^3)/abc
= [(a + b + c)^3 - 3ab(a + b + c) - 3bc(a + b + c) - 3ac(a + b + c) + 3abc ]/abc
= [(0)^3 - 3ab(0) - 3bc(0) - 3ac(0) + 3abc ]/abc
= 3abc/abc
= 3
Answer:
a²/bc + b²/ac + c²/ab = 3
Step-by-step explanation:
A+b+c=0 find a2/bc+b2/ac+c2/ab
a²/bc + b²/ac + c²/ab
= (1/abc) (a³ + b³ + c³)
using ( a + b)³ = a³ + b³ + 3ab(a+b)
=(1/abc) ( (a+b)³ - 3ab(a+b) + c³)
a + b + c = 0 => a+b = -c
Putting value of a +b = - c
= (1/abc) ( (-c)³ - 3ab(-c) + c³)
(-c)³ = - (c)³
= (1/abc) ( -c³ + 3abc + c³)
= (1/abc) ( 3abc)
Cancelling abc from numerator & denominator
= 3
Hence
a²/bc + b²/ac + c²/ab = 3