Question

If $\frac{a}{b}=\frac{b}{c}$ and $a,b,c$ > 0 then show that,

(i) (a + b + c) (b $-$ c) = ab $-$ c²

(ii) (a² + b²) (b² + c² ) = (ab + bc)²

(iii) $\frac{a^2+b^2}{ab}=\frac{a+c}{b}$

Answer 1

$\frac{a}{b}=\frac{b}{c}\Rightarrow b^2=ac.....\left(1\right)$

(i)
$$\begin{gathered} \left(a+b+c\right)\left(b-c\right) \\ =ab-ac+b^2-bc+bc-c^2 \\ =ab-ac+ac-c^2\left[\mathrm{Using}\left(1\right)\right] \\ =ab-c^2 \end{gathered}$$
(ii)
$$\begin{gathered} \left(a^2+b^2\right)\left(b^2+c^2\right) \\ =\left(a^2+ac\right)\left(ac+c^2\right)\left[\mathrm{Using}\left(1\right)\right] \\ =a\left(a+c\right)\times c\left(a+c\right) \\ =ac{\left(a+c\right)}^2 \\ =b^2{\left(a+c\right)}^2\left[\mathrm{Using}\left(1\right)\right] \\ ={\left(ab+bc\right)}^2 \end{gathered}$$
(iii)
$$\begin{gathered} \frac{a^2+b^2}{ab} \\ =\frac{{\displaystyle a^2+ac}}{{\displaystyle ab}}\left[\mathrm{Using}\left(1\right)\right] \\ =\frac{{\displaystyle a\left(a+c\right)}}{{\displaystyle ab}} \\ =\frac{a+c}{b} \end{gathered}$$