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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: $\left(5a-7b{\right)}^{3}+\left(9c-5a{\right)}^{3}+\left(7b-9c{\right)}^{3}$.

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Solution

## $\frac{\left(\sqrt{5}+\sqrt{3}\right)}{\left(\sqrt{5}-\sqrt{3}\right)}=a+\sqrt{15}b\phantom{\rule{0ex}{0ex}}\mathrm{Rationalising}\mathrm{the}\mathrm{denomiantor}\mathrm{of}\text{the}\mathrm{left}\mathrm{side},\text{we get:}\phantom{\rule{0ex}{0ex}}\frac{\left(\sqrt{5}+\sqrt{3}\right)}{\left(\sqrt{5}-\sqrt{3}\right)}×\frac{\left(\sqrt{5}+\sqrt{3}\right)}{\left(\sqrt{5}+\sqrt{3}\right)}=a+\sqrt{15}b\phantom{\rule{0ex}{0ex}}⇒\frac{{\left(\sqrt{5}+\sqrt{3}\right)}^{2}}{{\left(\sqrt{5}\right)}^{2}-{\left(\sqrt{3}\right)}^{2}}=a+\sqrt{15}b\phantom{\rule{0ex}{0ex}}⇒\frac{{\left(\sqrt{5}\right)}^{2}+{\left(\sqrt{3}\right)}^{2}+2\sqrt{5}×\sqrt{3}}{5-3}=a+\sqrt{15}b\phantom{\rule{0ex}{0ex}}⇒\frac{5+3+2\sqrt{15}}{2}=a+\sqrt{15}b\phantom{\rule{0ex}{0ex}}⇒\frac{8+2\sqrt{15}}{2}=a+\sqrt{15}b\phantom{\rule{0ex}{0ex}}⇒\frac{2\left(4+\sqrt{15}\right)}{2}=a+\sqrt{15}b\phantom{\rule{0ex}{0ex}}⇒4+\sqrt{15}=a+\sqrt{15}b\mathrm{Comparing}\mathrm{the}\mathrm{left}\mathrm{and}\mathrm{right}\mathrm{side}s,\mathrm{we}\mathrm{get}a=4,b=1.$ OR ${\left(5a-7b\right)}^{3}+{\left(9c-5a\right)}^{3}+{\left(7b-9c\right)}^{3}\phantom{\rule{0ex}{0ex}}Let\phantom{\rule{0ex}{0ex}}x=5a-7b\phantom{\rule{0ex}{0ex}}y=9c-5a\phantom{\rule{0ex}{0ex}}z=7b-9c\phantom{\rule{0ex}{0ex}}\therefore x+y+z=0\phantom{\rule{0ex}{0ex}}\text{If}x+y+z=0,\phantom{\rule{0ex}{0ex}}\text{t}\mathrm{hen}{x}^{3}+{y}^{3}+{z}^{3}=3xyz\phantom{\rule{0ex}{0ex}}\mathrm{Therefore},\phantom{\rule{0ex}{0ex}}{\left(5a-7b\right)}^{3}+{\left(9c-5a\right)}^{3}+{\left(7b-9c\right)}^{3}=3\left(5a-7b\right)\left(9c-5a\right)\left(7b-9c\right)\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}$

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