Question

if a3+b3+c3=3abc and a+b+c=0 show that (b+c)sq /3bc + (c+a)sq / 3ac + (a+b) sq / 3ab = 1

Answer 1

Algebra,

We have
a3+b3+c3=3abc....(1) and a+b+c=0.....(2)

Have to prove that,
(b+c)² /3bc + (c+a)² / 3ac + (a+b)² / 3ab = 1

Now from equation 1 we can write
a +b +c = 0
or, (a + b) = - c...(3)
similarly (b+c)=-a...(4)
and (c+a) = - b...(5)

LHS. = (b+c)² /3bc + (c+a)² / 3ac + (a+b)² / 3ab
= (-a)² /3bc + (-b)² / 3ac + (-c)² / 3ab

= a²/3bc + b²/3ac + c²/3ab [placing the values from the equations 3,4 and 5]

= a³/3abc + b³/3abc + c³/3abc

= (a³ + b³ + c³)/3abc

= (3abc)/3abc [from the equation 1]

= 1 = RHS [proved]

That's it
Hope it helped (｡ŏ﹏ŏ)

Answer 2

Answer:

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