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Question
If b is the mean proportion between a and c, show that: `(a^4 + a^2b^2 + b^4)/(b^4 + b^2c^2 + c^4) = a^2/c^2`.
Solution
Given, b is the mean proportion between a and c.
`=> a/b = b/c = k` ...(Say)
`=>` a = bk, b = ck
`=>` a = (ck)k = ck2, b = ck
L.H.S = `(a^4 + a^2b^2 + b^4)/(b^4 + b^2c^2 + c^4)`
= `((ck^2)^4 + (ck^2)^2 (ck)^2 + (ck)^4)/((ck)^4 + (ck)^2 c^2 + c^4)`
= `(c^4k^8 + (c^2k^4)(c^2k^2) + c^4k^4)/(c^4k^4 + (c^2k^2)c^2 + c^4)`
= `(c^4k^8 + c^4k^6 + c^4k^4)/(c^4k^4 + c^4k^2 + c^4)`
= `(c^4k^4(k^4 + k^2 + 1))/(c^4(k^4 + k^2 + 1))`
= k4
R.H.S = `a^2/c^2`
= `((ck^2)^2)/c^2`
= `(c^2k^4)/c^2`
= k4
Hence, L.H.S = R.H.S
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