i Hence proved ii Hence proved iii Hence proved

Selina Solutions Class 9 Mathematics Solutions for Exercise 1(D) in Chapter 1 - Chapter 1- Rational and Irrational Numbers

Question 10 Exercise 1(D)

Show that:

(i) $x^{3}+\frac{1}{x^{3}}=52, \text { if } x=2+\sqrt{3}$

(ii) $x^{2}+\frac{1}{x^{2}}=34, \text { if } x=3+2 \sqrt{2}$

(iii) $\frac{3 \sqrt{2}-2 \sqrt{3}}{3 \sqrt{2}+2 \sqrt{3}}+\frac{2 \sqrt{3}}{\sqrt{3}-\sqrt{2}}=11$

Answer:

(i) $\begin{aligned} &\text {We know that, }(a+b)^{3}=a^{3}+b^{3}+3 a b(a+b)\\ &x^{3}+\frac{1}{x^{3}}-(2+\sqrt{3})^{3}+\frac{1}{(2+\sqrt{3})^{3}}\\ &\text {Here}, \operatorname{taking}(2+\sqrt{3})^{3}\\ &\Rightarrow(2+\sqrt{3})^{3}=2^{3}+\sqrt{3}^{3}+3 \times 2 \times \sqrt{3}(2+\sqrt{3})\\ &=8+3 \sqrt{3}+6 \sqrt{3}(2+\sqrt{3})\\ &=8+3 \sqrt{3}+12 \sqrt{3}+18\\ &=26+15 \sqrt{3}\\ &\text { Now, }(2+\sqrt{3})^{3}+\frac{1}{(2+\sqrt{3})^{3}}=26+15 \sqrt{3}+\frac{1}{26+15 \sqrt{3}}\\ &\text {Taking} \frac{1}{26+15 \sqrt{3}}\\ &\Rightarrow \frac{1}{26+15 \sqrt{3}} \times \frac{26-15 \sqrt{3}}{26-15 \sqrt{3}}=\frac{26-15 \sqrt{3}}{676-675}=26-15 \sqrt{3}\\ &=26+15 \sqrt{3}+26-15 \sqrt{3}=52 \end{aligned}$

Hence proved.

(ii) $\begin{aligned} &\text {We know that,}(a+b)^{2}=a^{2}+b^{2}+2 a b\\ &x^{2}+\frac{1}{x^{2}}-(3+2 \sqrt{2})^{2}+\frac{1}{(3+2 \sqrt{2})^{2}}\\ &-(9+12 \sqrt{2}+8)+\frac{1}{(9+12 \sqrt{2}+8)}\\ &=(17+12 \sqrt{2})+\frac{1}{(17+12 \sqrt{2})}\\ &\operatorname{Taking} \frac{1}{(17+12 \sqrt{2})} \text { we get }\\ &\frac{1}{(17+12 \sqrt{2})} \times \frac{(17-12 \sqrt{2})}{(17-12 \sqrt{2})}=\frac{(17-12 \sqrt{2})}{289-288}=17-12 \sqrt{2}\\ &\therefore(17+12 \sqrt{2})+\frac{1}{(17+12 \sqrt{2})}-17+12 \sqrt{2}+17-12 \sqrt{2}-34 \end{aligned}$

Hence proved.

(iii) $\begin{array}{l} \frac{3 \sqrt{2}-2 \sqrt{3}}{3 \sqrt{2}+2 \sqrt{3}}+\frac{2 \sqrt{3}}{\sqrt{3}-\sqrt{2}} \\ \text { First, } \operatorname{taking} \frac{3 \sqrt{2}-2 \sqrt{3}}{3 \sqrt{2}+2 \sqrt{3}} \\ \frac{3 \sqrt{2}-2 \sqrt{3}}{3 \sqrt{2}+2 \sqrt{3}} \times \frac{3 \sqrt{2}-2 \sqrt{3}}{3 \sqrt{2}-2 \sqrt{3}}=\frac{(3 \sqrt{2}-2 \sqrt{3})^{2}}{18-12}=\frac{18-12 \sqrt{6}+12}{6} \\ \frac{-6(3-2 \sqrt{6}+2)}{6}=5-2 \sqrt{6} \\ \text { Now. taking } \frac{2 \sqrt{3}}{\sqrt{3}-\sqrt{2}} \\ \frac{2 \sqrt{3}}{\sqrt{3}-\sqrt{2}} \times \frac{\sqrt{3}+\sqrt{2}}{\sqrt{3}+\sqrt{2}}=\frac{6+2 \sqrt{6}}{3-2}=6+2 \sqrt{6} \\ \therefore \frac{3 \sqrt{2}-2 \sqrt{3}}{3 \sqrt{2}+2 \sqrt{3}}+\frac{2 \sqrt{3}}{\sqrt{3}-\sqrt{2}}=5-2 \sqrt{6}+6+2 \sqrt{6}=11 \end{array}$

Hence proved.

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