Question

If $\frac{\left(\sqrt{5}+\sqrt{3}\right)}{\left(\sqrt{5}-\sqrt{3}\right)}=a+\sqrt{15}b$, find the values of a and b.
OR
Factorise: $(5a-7b{)}^3+(9c-5a{)}^3+(7b-9c{)}^3$.

Answer 1

$$\begin{gathered} \frac{\left(\sqrt{5}+\sqrt{3}\right)}{\left(\sqrt{5}-\sqrt{3}\right)}=a+\sqrt{15}b \\ \mathrm{Rationalising}\mathrm{the}\mathrm{denomiantor}\mathrm{of}\text{the}\mathrm{left}\mathrm{side},\text{we get:} \\ \frac{\left(\sqrt{5}+\sqrt{3}\right)}{\left(\sqrt{5}-\sqrt{3}\right)}\times \frac{\left(\sqrt{5}+\sqrt{3}\right)}{\left(\sqrt{5}+\sqrt{3}\right)}=a+\sqrt{15}b \\ \Rightarrow \frac{{\left(\sqrt{5}+\sqrt{3}\right)}^2}{{\left(\sqrt{5}\right)}^2-{\left(\sqrt{3}\right)}^2}=a+\sqrt{15}b \\ \Rightarrow \frac{{\left(\sqrt{5}\right)}^2+{\left(\sqrt{3}\right)}^2+2\sqrt{5}\times \sqrt{3}}{5-3}=a+\sqrt{15}b \\ \Rightarrow \frac{5+3+2\sqrt{15}}{2}=a+\sqrt{15}b \\ \Rightarrow \frac{8+2\sqrt{15}}{2}=a+\sqrt{15}b \\ \Rightarrow \frac{2\left(4+\sqrt{15}\right)}{2}=a+\sqrt{15}b \\ \Rightarrow 4+\sqrt{15}=a+\sqrt{15}b\mathrm{Comparing}\mathrm{the}\mathrm{left}\mathrm{and}\mathrm{right}\mathrm{side}s,\mathrm{we}\mathrm{get}a=4,b=1. \end{gathered}$$

OR

$$\begin{gathered} {\left(5a-7b\right)}^3+{\left(9c-5a\right)}^3+{\left(7b-9c\right)}^3 \\ Let \\ x=5a-7b \\ y=9c-5a \\ z=7b-9c \\ \therefore x+y+z=0 \\ \text{If}x+y+z=0, \\ \text{t}\mathrm{hen}{x}^3+y^3+z^3=3xyz \\ \mathrm{Therefore}, \\ {\left(5a-7b\right)}^3+{\left(9c-5a\right)}^3+{\left(7b-9c\right)}^3=3\left(5a-7b\right)\left(9c-5a\right)\left(7b-9c\right) \end{gathered}$$