If A⅓ + b⅔ + c =0 then show that (a + b + c³)³ = 27ab²c³
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Step-by-step explanation:
a⅓ + b⅔ + c = 0
=> (a⅓ + b⅔ + c)³ = 0³
=> (a⅓)³ + (b⅔)³ + c³ - 3(a⅓ * b⅔ * c) = 0
=> (a⅓)³ + (b⅔)³ + c³ = 3a⅓ b⅔ c
=> a+b²+c³ = 3a⅓ b⅔ c
on cubing both sides,
=> (a+b²+c³)³ = (3a⅓ b⅔ c)³
=> (a+b²+c³)³ = 3³(a⅓ )³(b⅔)³ c³
=> (a+b²+c³)³ = 27ab²c³ , which is the required equation
Hence proved
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