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`.

Answer 1

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