Question

Rationalising factor of cube root of 25 + (1/ cube root of 25 -1)

Answer 1

GIVEN :

The expression is $\frac{\sqrt[3]{25}+1}{\sqrt[3]{25}-1}$

TO RATIONALISE :

The  given expression $\frac{\sqrt[3]{25}+1}{\sqrt[3]{25}-1}$

SOLUTION :

Given expression is $\frac{\sqrt[3]{25}+1}{\sqrt[3]{25}-1}$

Now solving the given expression as below :

$\frac{\sqrt[3]{25}+1}{\sqrt[3]{25}-1}$

$=\frac{5\sqrt{5}+1}{5\sqrt{5}-1}$

Multiplying and dividing by $5\sqrt{5}+1$  by the denominator's conjugate.

$=\frac{5\sqrt{5}+1}{5\sqrt{5}-1}\times \frac{5\sqrt{5}+1}{5\sqrt{5}+1}$

By using the Algebraic Identity :

$(a+b)(a-b)=a^2-b^2$

$=\frac{(5\sqrt{5}+1)^2}{(5\sqrt{5})^2-1^2}$

By using the Algebraic Identity :

$(a+b)^2=a^2+2ab+b^2$

$=\frac{(5\sqrt{5})^2+2(5\sqrt{5})(1)+1^2}{5^2(\sqrt{5})^2-1^2}$

$=\frac{5^2\sqrt{5}^2+10\sqrt{5}+1}{25(5)-1}$

By using the square root property

$\sqrt{x}^2=x$

$=\frac{25(5)+10\sqrt{5}+1}{125-1}$

$=\frac{125+10\sqrt{5}+1}{124}$

By adding the like terms

$=\frac{126+10\sqrt{5}}{124}$

After simplifying the terms,

$=\frac{63+5\sqrt{5}}{62}$

$\frac{\sqrt[3]{25}+1}{\sqrt[3]{25}-1}=\frac{63+5\sqrt{5}}{62}$

∴ the given expression $\frac{\sqrt[3]{25}+1}{\sqrt[3]{25}-1}$ rationalised into $\frac{63+5\sqrt{5}}{62}$.