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)`

Answer 1

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.