$\sqrt[3]{a}+\sqrt[3]{b}+\sqrt[3]{c}=0\backslash sqrt[3]a+\backslash sqrt[3]b+\; \backslash sqrt[3]c=0$ then $(a+b+c{)}^3=\mathrm{.}\mathrm{.}\mathrm{.}\mathrm{.}\mathrm{.}\mathrm{.}\mathrm{.}\mathrm{.}\mathrm{.}\mathrm{.}\mathrm{.}(a+b+c)^3\; =\; ...........$

Rationalising factor of $\sqrt[3]{16}-\sqrt[3]{4}+1\backslash sqrt[3]\{16\}-\backslash sqrt[3]4+1$ 

$\frac{l}{\sqrt{10}+\sqrt{14}+\sqrt{15}+\sqrt{21}}=\frac{\sqrt{10}-\sqrt{14}-\sqrt{1}5+\sqrt{2}1}{k}\backslash frac\{l\}\{\backslash sqrt\{10\}+\backslash sqrt\{14\}+\backslash sqrt\{15\}+\backslash sqrt\{21\}\}=\backslash frac\{\backslash sqrt\{10\}-\backslash sqrt\{14\}-\backslash sqrt15+\backslash sqrt21\}\{k\}$ then k = …………

A rationalising factor of $\sqrt[3]{25}+\frac{1}{\sqrt[3]{25}}-1\backslash sqrt[3]\{25\}+\backslash frac\{1\}\{\backslash sqrt[3]\{25\}\}-1$ is ……………

$\frac{\sqrt{x^2+1}+\sqrt{x^2-1}}{\sqrt{x^2+1}-\sqrt{x^2-1}}+\frac{\sqrt{x^2+1}-\sqrt{x^2-1}}{\sqrt{x^2+1}+\sqrt{x^2-1}}=\mathrm{.}\mathrm{.}\mathrm{.}\mathrm{.}\mathrm{.}\mathrm{.}\mathrm{.}\mathrm{.}\mathrm{.}\mathrm{.}\backslash frac\{\backslash sqrt\{x^2+1\}+\backslash sqrt\{x^2-1\}\}\{\backslash sqrt\{x^2+1\}-\backslash sqrt\{x^2-1\}\}+\backslash frac\{\backslash sqrt\{x^2+1\}-\backslash sqrt\{x^2-1\}\}\{\backslash sqrt\{x^2+1\}+\backslash sqrt\{x^2-1\}\}=..........$

Arithmetic mean of two surds is $5+9\sqrt{2}5+9\backslash sqrt2$ and one of them is $1+12\sqrt{2}1+12\backslash sqrt2$ then second surd is …………

$\frac{1+\sqrt{2}}{3-2\sqrt{2}}=a+b\sqrt{2}\backslash frac\{1+\backslash sqrt2\}\{3-2\backslash sqrt2\}=a+b\backslash sqrt2$ then (a, b) = ……………

$x=2\sqrt{2}+\sqrt{7}x=2\backslash sqrt2+\backslash sqrt7$ then $\frac{1}{2}(x-\frac{1}{x})\backslash frac\{1\}\{2\}(x-\backslash frac\{1\}\{x\})$ = …………

If the product of two irrational numbers is rational, then which of the following can be concluded ?

Which of the following is always true?