Question

solve 4.(iii) and (iv)



please​

Answer 1

NOTE :

$\bigstar \sqrt{a}\times \sqrt{b}=\sqrt{ab}\\\\\bigstar \mathsf{(a+b)(a-b)=a^2-b^2}\\\\\bigstar \mathsf{(a+b)^3=a^3+b^3+3ab(a+b)}\\\\\bigstar \mathsf{(a-b)^3=a^3-b^3-3ab(a-b)}\\\\\bigstar \mathsf{(a+b)^2=a^2+b^2+2ab}$

4 (iii)

Given :-

$x=\sqrt{3}-\sqrt{2}\\\\\implies x^3=(\sqrt{3}-\sqrt{2})^3\\\\\implies x^3=(\sqrt{3})^3-(\sqrt{2})^3-3\sqrt{3}\times \sqrt{2}(\sqrt{3}-\sqrt{2})\\\\\implies x^3=(\sqrt{3})^3-(\sqrt{2})^3-3\sqrt{6}(\sqrt{3}-\sqrt{2})$

$\dfrac{1}{x}=\dfrac{1}{\sqrt{3}-\sqrt{2}}\\\\\implies \dfrac{1}{x}=\dfrac{1}{\sqrt{3}-\sqrt{2}}\times \dfrac{\sqrt{3}+\sqrt{2}}{\sqrt{3}+\sqrt{2}}\\\implies \dfrac{1}{x}=\dfrac{\sqrt{3}+\sqrt{2}}{3-2}\\\\\implies \dfrac{1}{x^3}=(\sqrt{3}+\sqrt{2})^3\\\\\implies \dfrac{1}{x^3}=(\sqrt{3})^3+(\sqrt{2})^3+3\sqrt{3}\times \sqrt{2}(\sqrt{3}+\sqrt{2})\\\\\implies \dfrac{1}{x^3}=(\sqrt{3})^3+(\sqrt{2})^3+3\sqrt{6}(\sqrt{3}+\sqrt{2})$

$x^3+\dfrac{1}{x^3}=[(\sqrt{3})^3-(\sqrt{2})^3-3\sqrt{6}(\sqrt{3}-\sqrt{2})]+[(\sqrt{3})^3+(\sqrt{2})^3+3\sqrt{6}(\sqrt{3}+\sqrt{2})]\\\\\implies 2(\sqrt{3})^3+3\sqrt{6}\times \sqrt{2}+3\sqrt{6}\times \sqrt{2}\\\\\implies 2(\sqrt{3})^3+6\sqrt{12}\\\\\bf{Note\:\sqrt{3}^3=(\sqrt{3})^2\times \sqrt{3}=3\sqrt{3}}\\\\\implies \mathsf{6\sqrt{3}+6\sqrt{3\times 4}}\\\\\implies \mathsf{6\sqrt{3}+12\sqrt{3}}\\\\\implies \boxed{\mathsf{18\sqrt{3}}}$

4 (iv)

We already found out that :-

$x=\sqrt{3}-\sqrt{2}\\\\\dfrac{1}{x}=\sqrt{3}+\sqrt{2}\\\\x^3+\dfrac{1}{x^3}=18\sqrt{3}$

$x^2+\dfrac{1}{x^2}\\\\\implies \mathsf{(\sqrt{3}-\sqrt{2})^2+(\sqrt{3}+\sqrt{2})^2}\\\\\implies \mathsf{3+2-2\sqrt{6}+3+2+2\sqrt{6}}\\\\\implies 10$

We have to find the value of :

$x^3+\dfrac{1}{x^3}-3(x^2+\dfrac{1}{x^2})+x+\dfrac{1}{x}\\\\\implies \mathsf{18\sqrt{3}-3\times 10+\sqrt{3}-\sqrt{2}+\sqrt{3}+\sqrt{2}}\\\\\implies \mathsf{18\sqrt{3}-30+2\sqrt{3}}\\\implies \mathsf{\sqrt{3}(18+2)-30}\\\\\implies \boxed{\mathsf{20\sqrt{3}-30}}$