if a+b+c=0 then find a4+b4+c4/a2b2+b2c2+c2a2
Lipimishra2:
Answer is 2?
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143
Well, quite a lengthy question with a simple solution.
a+b+c = 0
(a+b)² = (-c)²
a²+b²+2ab = c²
(a²+b²-c²)² = (-2ab)²
a⁴+b⁴+c⁴+2a²b²-2b²c²-2c²a² = 4a²b²
a⁴+b⁴+c⁴+2a²b²-4a²b²-2b²c²-2c²a² = 0
a⁴+b⁴+c⁴-2a²b²-2b²c²-2c²a²= 0
a⁴+b⁴+c⁴= 2a²b²+2b²c²+2c²a²
a⁴+b⁴+c⁴= 2(a²b²+b²c²+c²a²)
Substituting this value with a⁴+b⁴+c⁴/a²b²+b²c²+c²a²
We get,
2(a²b²+b²c²+c²a²)/ a²b²+b²c²+c²a² = 2
a+b+c = 0
(a+b)² = (-c)²
a²+b²+2ab = c²
(a²+b²-c²)² = (-2ab)²
a⁴+b⁴+c⁴+2a²b²-2b²c²-2c²a² = 4a²b²
a⁴+b⁴+c⁴+2a²b²-4a²b²-2b²c²-2c²a² = 0
a⁴+b⁴+c⁴-2a²b²-2b²c²-2c²a²= 0
a⁴+b⁴+c⁴= 2a²b²+2b²c²+2c²a²
a⁴+b⁴+c⁴= 2(a²b²+b²c²+c²a²)
Substituting this value with a⁴+b⁴+c⁴/a²b²+b²c²+c²a²
We get,
2(a²b²+b²c²+c²a²)/ a²b²+b²c²+c²a² = 2
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