Question

Find the rationalising factor of the given binomial surd: $2^{\frac{1}{3}}+2^{\frac{1}{3}}$

Answer 1

The given binomial surd is $2^{\frac{1}{3}}+2^{-\frac{1}{3}}$.

Let $a=2^{\frac{1}{3}}$ and $b=2^{-\frac{1}{3}}$, therefore,

$a=2^{\frac{1}{3}}\Rightarrow a^3={\left(2^{\frac{1}{3}}\right)}^3\Rightarrow a^3=2^{\frac{3}{3}}\Rightarrow a^3=2$

Also,

$b=2^{-\frac{1}{3}}\Rightarrow b^3={\left(2^{-\frac{1}{3}}\right)}^3\Rightarrow b^3=2^{-\frac{3}{3}}\Rightarrow b^3=2^{-1}\Rightarrow b^3=\frac{1}{2}$

Thus, $a^3+b^3=2+\frac{1}{2}=\frac{4+1}{2}=\frac{5}{2}$

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

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

Since $\frac{5}{2}$ is a rational number.

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