Question
If $\frac{a}{b}=\frac{b}{c}$ and $a, b, c>0$, then show that, : $\left(a^2+b^2\right)\left(b^2+c^2\right)=(a b+b e)^2$
✓
Answer
$\text { Let } \frac{a}{b}=\frac{b}{c}=k$
$\therefore b=c k$
$\therefore a=b k=(c k) k$
$\therefore a=c k^2 \ldots \text { (ii) }$
$(a^2 + b^2)(b^2 + c^2) = (ab + bc)^2$^
$b = ck; a = ck^2$
$L.H.S = (a^2 + b^2) (b^2 + c^2)$
$= [(ck^2) + (ck)^2] [(ck)^2 + c^2] … [From (i) and (ii)]$
$= [c^2k^4 + c^2k^2] [c^2k^2 + c^2]$
$= c^2k^2 (k^2 + 1) c^2 (k^2 + 1)$
$= c4k^2 (k^2 + 1)^2$
$R.H.S = (ab + bc)^2$
$= [(ck^2) (ck) + (ck)c]^2 …[From (i) and (ii)]$
$= [c^2k^3 + c^2k]^2$
$= [c^2k (k^2 + 1)]2 = c^4(k^2 + 1)^2$
$\therefore L.H.S = R.H.S$
$\therefore (a^2 + b^2) (b^2 + c^2) = (ab + bc)^2$
$\therefore b=c k$
$\therefore a=b k=(c k) k$
$\therefore a=c k^2 \ldots \text { (ii) }$
$(a^2 + b^2)(b^2 + c^2) = (ab + bc)^2$^
$b = ck; a = ck^2$
$L.H.S = (a^2 + b^2) (b^2 + c^2)$
$= [(ck^2) + (ck)^2] [(ck)^2 + c^2] … [From (i) and (ii)]$
$= [c^2k^4 + c^2k^2] [c^2k^2 + c^2]$
$= c^2k^2 (k^2 + 1) c^2 (k^2 + 1)$
$= c4k^2 (k^2 + 1)^2$
$R.H.S = (ab + bc)^2$
$= [(ck^2) (ck) + (ck)c]^2 …[From (i) and (ii)]$
$= [c^2k^3 + c^2k]^2$
$= [c^2k (k^2 + 1)]2 = c^4(k^2 + 1)^2$
$\therefore L.H.S = R.H.S$
$\therefore (a^2 + b^2) (b^2 + c^2) = (ab + bc)^2$
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