Question

If $$b$$ is the mean proportion between $$a$$ and $$c$$, show that $$\frac{a^4 + a^2b^2 + b^4}{b^4 + b^2c^2 + c^4} = \frac{a^2}{c^2}$$.

[Hint : Substitute \[b^{2} = ac\] in L.H.S. to get R.H.S.]

Answer 1

Given: $$b$$ is the mean proportion between $$a$$ and $$c$$, so $$b^2 = ac$$.

To prove: $$\frac{a^4 + a^2b^2 + b^4}{b^4 + b^2c^2 + c^4} = \frac{a^2}{c^2}$$

Proof:

1. Since $$b^2 = ac$$, we have $$b^4 = (b^2)^2 = a^2 c^2$$.
2. Substitute $$b^2 = ac$$ and $$b^4 = a^2 c^2$$ into the numerator: $$a^4 + a^2(ac) + a^2 c^2 = a^4 + a^3 c + a^2 c^2 = a^2(a^2 + ac + c^2)$$
3. Substitute $$b^2 = ac$$ and $$b^4 = a^2 c^2$$ into the denominator: $$a^2 c^2 + (ac)c^2 + c^4 = a^2 c^2 + ac^3 + c^4 = c^2(a^2 + ac + c^2)$$
4. $$\text{L.H.S.} = \frac{a^2(a^2 + ac + c^2)}{c^2(a^2 + ac + c^2)} = \frac{a^2}{c^2}$$
5. $$\text{L.H.S.} = \text{R.H.S.}$$

Hence proved.