Question

If a, b, c are in G.P., prove that:

(i) a (b² + c²) = c (a² + b²)

(ii) $a^2{b}^2{c}^2\left(\frac{1}{a^3}+\frac{1}{b^3}+\frac{1}{c^3}\right)=a^3+b^3+c^3$

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

(iv) $\frac{1}{a^2-b^2}+\frac{1}{b^2}=\frac{1}{b^2-c^2}$

(v) (a + 2b + 2c) (a − 2b + 2c) = a² + 4c².

Answer 1

a, b and c are in G.P.
$\therefore b^2=ac$ .......(1)

$$\begin{gathered} \left(\mathrm{i}\right)\mathrm{LHS}=a\left(b^2+c^2\right) \\ =a{b}^2+a{c}^2 \\ =a\left(ac\right)+c\left(b^2\right)\left[\mathrm{Using}\left(1\right)\right] \\ =c\left(a^2+b^2\right)=\mathrm{RHS} \end{gathered}$$

$$\begin{gathered} \left(\mathrm{ii}\right)\mathrm{LHS}=a^2{b}^2{c}^2\left(\frac{1}{a^3}+\frac{1}{b^3}+\frac{1}{c^3}\right) \\ =\frac{b^2{c}^2}{a}+\frac{a^2{c}^2}{b}+\frac{a^2{b}^2}{c} \\ =\frac{\left(ac\right)c^2}{a}+\frac{{\left(b^2\right)}^2}{b}+\frac{a^2\left(ac\right)}{c}\left[\mathrm{Using}\left(1\right)\right] \\ =a^3+b^3+c^3=\mathrm{RHS} \end{gathered}$$

$$\begin{gathered} \left(\mathrm{iii}\right)\mathrm{LHS}=\frac{{\left(a+b+c\right)}^2}{a^2+b^2+c^2} \\ =\frac{{\left(a+b+c\right)}^2}{a^2-b^2+c^2+2b^2} \\ =\frac{{\left(a+b+c\right)}^2}{a^2-b^2+c^2+2ac}\left[\mathrm{Using}\left(1\right)\right] \\ \mathbf{=}\frac{{\left(a+b+c\right)}^2}{\left(a+b+c\right)\left(a-b+c\right)}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\left[\because \left(\mathrm{a}+\mathrm{b}+\mathrm{c}\right)\left(\mathrm{a}-\mathrm{b}+\mathrm{c}\right)={\mathrm{a}}^2-{\mathrm{b}}^2+{\mathrm{c}}^2+2\mathrm{ac}\right] \\ =\frac{\left(a+b+c\right)}{\left(a-b+c\right)}=\mathrm{RHS} \end{gathered}$$

$$\begin{gathered} \left(\mathrm{iv}\right)\mathrm{LHS}=\frac{1}{a^2-b^2}+\frac{1}{b^2} \\ =\frac{b^2+a^2-b^2}{\left(a^2-b^2\right)b^2} \\ =\frac{a^2}{\left(a^2{b}^2-b^4\right)} \\ =\frac{a^2}{a^2\left(ac\right)-{\left(ac\right)}^2} \\ =\frac{1}{ac-c^2} \\ =\frac{1}{b^2-c^2}=\mathrm{RHS} \end{gathered}$$

$$\begin{gathered} \left(\mathrm{v}\right)\mathrm{LHS}=\left(a+2b+2c\right)\left(a-2b+2c\right) \\ =a^2-4b^2+4c^2+4ac \\ =a^2-4ac+4c^2+4ac\left[\mathrm{Using}\left(1\right)\right] \\ =a^2+4c^2=\mathrm{RHS} \end{gathered}$$