Question

Find the factor which will rationalise $\sqrt{3}+\sqrt[3]{35}$.

Answer 1

Let $x=3^{\frac{1}{2}},y=5^{\frac{1}{3}}$; then $x^6$ and $y^6$ are both rational, and
$x^6-y^6=(x+y)(x^5-x^{4}y+x^3{y}^2+x^2{y}^3-x{y}^4-y^5)$;
Thus, substituting for $x$ and $y$ the required factor is
$3^{\frac{5}{2}}-3^{\frac{4}{2}}.5^{\frac{1}{3}}+3^{\frac{3}{2}}.5^{\frac{2}{3}}-3^{\frac{2}{2}}.5^{\frac{3}{3}}+3^{\frac{1}{2}}.5^{\frac{4}{3}}-5^{\frac{5}{3}}$,
or $3^{\frac{5}{2}}-9.5^{\frac{1}{3}}+3^{\frac{3}{2}}.5^{\frac{2}{3}}-15+3^{\frac{1}{2}}.5^{\frac{4}{3}}-5^{\frac{5}{3}}$;
and the rational product is $3^{\frac{6}{2}}-5^{\frac{6}{3}}=3^3-5^2=2$