Question
If $b$ is the mean proportion between $a$ and $c,$ show that:
$\frac{a^4+a^2 b^2+b^4}{b^4+b^2 c^2+c^4}=\frac{a^2}{c^2}$
$\frac{a^4+a^2 b^2+b^4}{b^4+b^2 c^2+c^4}=\frac{a^2}{c^2}$
✓
Answer
Given that $b$ is the mean proportion between $a$ and $c.$
$\frac{ a }{ b }=\frac{ b }{ c }= k$
$\Rightarrow b = ck ; a = bk =( ck ) k = ck ^2$
$L.H.S =\frac{a^4+a^2 b^2+b^4}{b^4+b^2 c^2+c^4}$
$=\frac{\left( ck ^2\right)^4+\left( ck ^2\right)^2( ck )^2+( ck )^4}{( ck )^4+(c k)^2 c ^2+ c ^4}$
$=\frac{ c ^4 k ^8+ c ^4 k ^6+ c ^4 k ^4}{ c ^4 k ^4+ c ^4 k ^2+ c ^4}$
$=\frac{ c ^4 k ^4\left( k ^4+ k ^2+1\right)}{ c ^4\left( k ^4+ k ^2+1\right)}$
$= k ^4 \ldots \ldots \ldots . \text { (i) }$
$R.H.S =\frac{ a ^2}{ c ^2}=\frac{\left( ck ^2\right)^2}{ c ^2}=\frac{ c ^2 k ^4}{ c ^2}= k ^4 \ldots \ldots. (ii)$
From $(i)$ and $(ii)$ we get
$L.H.S = R.H.S$
$\Rightarrow \frac{ a ^4+ a ^2 b ^2+ b ^4}{ b ^4+ b ^2 c ^2+ c ^4}=\frac{ a ^2}{ c ^2}$
Hence proved.
$\frac{ a }{ b }=\frac{ b }{ c }= k$
$\Rightarrow b = ck ; a = bk =( ck ) k = ck ^2$
$L.H.S =\frac{a^4+a^2 b^2+b^4}{b^4+b^2 c^2+c^4}$
$=\frac{\left( ck ^2\right)^4+\left( ck ^2\right)^2( ck )^2+( ck )^4}{( ck )^4+(c k)^2 c ^2+ c ^4}$
$=\frac{ c ^4 k ^8+ c ^4 k ^6+ c ^4 k ^4}{ c ^4 k ^4+ c ^4 k ^2+ c ^4}$
$=\frac{ c ^4 k ^4\left( k ^4+ k ^2+1\right)}{ c ^4\left( k ^4+ k ^2+1\right)}$
$= k ^4 \ldots \ldots \ldots . \text { (i) }$
$R.H.S =\frac{ a ^2}{ c ^2}=\frac{\left( ck ^2\right)^2}{ c ^2}=\frac{ c ^2 k ^4}{ c ^2}= k ^4 \ldots \ldots. (ii)$
From $(i)$ and $(ii)$ we get
$L.H.S = R.H.S$
$\Rightarrow \frac{ a ^4+ a ^2 b ^2+ b ^4}{ b ^4+ b ^2 c ^2+ c ^4}=\frac{ a ^2}{ c ^2}$
Hence proved.
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