Question

If x = \sqrt[3]{25}+\sqrt[3]{20}+\sqrt[3]{16} ,

find the value of x -\dfrac{1}{x^2}

Collected in the board: Cube root

Steven Zheng posted 25 minutes ago

Answer

x = \sqrt[3]{25}+\sqrt[3]{20}+\sqrt[3]{16}

= \sqrt[3]{5^2}+\sqrt[3]{4\cdot 5 }+\sqrt[3]{4^2}

Let

a = \sqrt[3]{5}

b=\sqrt[3]{4}

Therefore

x = a^2+ab+b^2

Using the difference of cubes formula

a^3-b^3 = (a-b)(a^2+ab+b^2)

5-4 = (a-b)x = 1

x=\dfrac{1}{a-b}

Find the reciprocal of x

\dfrac{1}{x} = a-b

Take square of \dfrac{1}{x} and substitute a^2+ab+b^2 back to x

\dfrac{1}{x^2} = a^2-2ab+b^2 =a^2+ab+b^2-3ab = x-3ab

Then,

x-\dfrac{1}{x^2} = x-(x-3ab) =3ab = 3\sqrt[3]{20}


Steven Zheng posted 25 minutes ago

Scroll to Top