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Question
If a, b, c are in continued proportion, prove that: `(1)/a^3 + (1)/b^3 + (1)/c^3 = a/(b^2c^2) + b/(c^2a^2) + c/(a^2b^2)`
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Solution
As a, b, c, are in continued proportion
Let `a/b = b/c` = k
L.H.S. = `(1)/a^3 + (1)/b^3 + (1)/c^3`
= `(1)/(ck^2)^3 + (1)/(ck)^3 + (1)/c^3`
= `(1)/(c^3k^6) + (1)/(c^3k^3) + (1)/c^3`
= `(1)/c^3[1/k^6 + 1/k^3 + 1/1]`
R.H.S. = `a/(b^2c^2) + b/(c^2a^2) + c/(a^2b^2)`
= `ck^2/((ck)^2c^2) + "ck"/(c^2(ck^2)^2) + c/((ck^2)^2(ck)^2)`
= `(ck^2)/(c^4k^2) + "ck"/(c^4k^4) + c/(c^4k^6)`
= `(1)/c^3 + (1)/(c^3k^3) + (1)/(c^3k^6)`
= `(1)/c^3[1 + 1/k^3 + 1/k^6]`
= `(1)/c^3[1/k^6 + 1/k^3 + 1]`
∴ L.H.S. = R.H.S.
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