Question

a3+b3+c3/a2b2c2=1/a3+1/b3+1/c3 show that a,b,c they are in continued proportion​

Answer 1

we have to prove that a, b and c are in continued proportion where (a³ + b³ + c³)/a²b²c² = 1/a³ + 1/b³ + 1/c³

Solution : here (a³ + b³ + c³)/a²b²c² = 1/a³ + 1/b³ + 1/c³

⇒a³/a²b²c² + b³/a²b²c² + c³/a²b²c² = 1/a³ + 1/b³ + 1/c³

⇒a/b²c² + b/c²a² + c/a²b² = 1/a³ + 1/b³ + 1/c³

⇒a/b²c² - 1/b³ + b/c²a² - 1/c³ + c/a²b² - 1/a³ = 0

⇒(ab - c²)/b³c² + (bc - a²)/c³a² + (ca - b²)/a³b² = 0

Now this will be true only when,

(ab - c²) = 0 ⇒a/c = c/b......(1)

(bc - a²) = 0 ⇒b/a = a/c .....(2)

(ca - b²) = 0 ⇒c/b = b/a .....(3)

From equations (1), (2) and (3) we get,

a/c = c/b = b/a

Therefore it is clear that a, b and c are in continued proportion.

Answer 2

Answer:

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