Question

Find the rationalising factor of the given binomial surd: $5^{\frac{1}{3}}+5^{-\frac{1}{3}}$

Answer 1

The given binomial surd is $5^{\frac{1}{3}}+5^{-\frac{1}{3}}$.

Let $a=5^{\frac{1}{3}}$ and $b=5^{-\frac{1}{3}}$, therefore,

$a=5^{\frac{1}{3}}\Rightarrow a^3={\left(5^{\frac{1}{3}}\right)}^3\Rightarrow a^3=5^{\frac{3}{3}}\Rightarrow a^3=5$

Also,

$b=5^{-\frac{1}{3}}\Rightarrow b^3={\left(5^{-\frac{1}{3}}\right)}^3\Rightarrow b^3=5^{-\frac{3}{3}}\Rightarrow b^3=5^{-1}\Rightarrow b^3=\frac{1}{5}$

Thus, $a^3+b^3=5+\frac{1}{5}=\frac{25+1}{5}=\frac{26}{5}$

But we also know that the formula for $a^3+b^3=(a+b)(a^2-ab+b^2)$, therefore,

$a^3+b^3=(a+b)(a^2-ab+b^2)\Rightarrow \frac{26}{5}=(5^{\frac{1}{3}}+5^{-\frac{1}{3}})[{\left(5^{\frac{1}{3}}\right)}^2-(5^{\frac{1}{3}}\times 5^{-\frac{1}{3}})+{\left(5^{-\frac{1}{3}}\right)}^2]\Rightarrow \frac{26}{5}=(5^{\frac{1}{3}}+5^{-\frac{1}{3}})[\left(5^{\frac{2}{3}}\right)-\left(5^{\frac{1}{3}-\frac{1}{3}}\right)+\left(5^{-\frac{2}{3}}\right)]\Rightarrow \frac{26}{5}=(5^{\frac{1}{3}}+5^{-\frac{1}{3}})[5^{\frac{2}{3}}+5^{-\frac{2}{3}}-\left(5^0\right)]\Rightarrow \frac{26}{5}=(5^{\frac{1}{3}}+5^{-\frac{1}{3}})(5^{\frac{2}{3}}+5^{-\frac{2}{3}}-1)$

Since $\frac{26}{5}$ is a rational number.

Hence, $(5^{\frac{2}{3}}+5^{-\frac{2}{3}}-1)$ is the rationalising factor of $(5^{\frac{1}{3}}+5^{-\frac{1}{3}})$.