An algebra problem by Mohammed Rizqallah

Algebra Level 3

7 + 50 2 3 + 7 − 50 2 3 = ? \sqrt[3]{7+\sqrt[2]{50}} + \sqrt[3]{7-\sqrt[2]{50}} = \, ?


Note: Treat − X 3 = X 3 \sqrt[3]{-X} = \sqrt[3]{X} for all real X X .


The answer is 2.

Mohammed Rizqallah by Mohammed Rizqallah , 23, Jordan

This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try refreshing the page, (b) enabling javascript if it is disabled on your browser and, finally, (c) loading the non-javascript version of this page . We're sorry about the hassle.

1 solution

X = 7 + 50 2 3 + 7 − 50 2 3 \large X = \sqrt[3]{7+\sqrt[2]{50}} + \sqrt[3]{7-\sqrt[2]{50}}

by taking the cube for each side X 3 = ( 7 + 50 2 3 + 7 − 50 2 3 ) 3 \large X^3 = (\sqrt[3]{7+\sqrt[2]{50}} + \sqrt[3]{7-\sqrt[2]{50}})^3

note : ( A + B ) 3 = A 3 + 3 ∗ A 2 ∗ B + 3 ∗ A ∗ B 2 + B 3 \large (A+B)^3 = A^3 + 3*A^2*B + 3*A*B^2 + B^3

So :

X 3 = ( 7 + 50 2 3 3 + 3 ∗ 7 + 50 2 3 2 ∗ 7 − 50 2 3 + 3 ∗ 7 + 50 2 3 ∗ 7 + 50 2 3 2 + 7 − 50 2 3 3 \large X^3 = ( \sqrt[3]{7+\sqrt[2]{50}}^3 + 3*\sqrt[3]{7+\sqrt[2]{50}}^2 * \sqrt[3]{7-\sqrt[2]{50}} + 3*\sqrt[3]{7+\sqrt[2]{50}} *\sqrt[3]{7+\sqrt[2]{50}}^2 +\sqrt[3]{7-\sqrt[2]{50}}^3

simplification it .. Then :

X 3 = ( 14 + 3 ∗ 7 + 50 2 3 2 ∗ 7 − 50 2 3 + 3 ∗ 7 + 50 2 3 ∗ 7 − 50 2 3 2 \large X^3 = (14 + 3*\sqrt[3]{7+\sqrt[2]{50}}^2 * \sqrt[3]{7-\sqrt[2]{50}} +3*\sqrt[3]{7+\sqrt[2]{50}} * \sqrt[3]{7-\sqrt[2]{50}}^2

Look carefully here :

X 3 = 14 + 3 ∗ ( ( 7 + 50 ) 2 ) ∗ ( 7 + 50 2 ) ∗ ( 7 − 50 2 ) 3 + 3 ∗ ( 7 + 50 ) 2 ∗ ( 7 + 50 2 ) ∗ ( 7 − 50 2 ) 3 \large X^3 =14+ 3 * (\sqrt[3]{(7+\sqrt[2]{50)}) * (7+\sqrt[2]{50}) * (7-\sqrt[2]{50}) } + 3*\sqrt[3]{(7+\sqrt[2]{50)} *(7+\sqrt[2]{50}) * (7-\sqrt[2]{50})}

( 7 + 50 2 ) ∗ ( 7 − 50 2 ) = − 1 (7+\sqrt[2]{50}) *(7- \sqrt[2]{50} )= -1

( 7 + 50 ) 2 ∗ − 1 3 + 3 ∗ ( 7 + 50 ) 2 ∗ − 1 3 \sqrt[3]{(7+\sqrt[2]{50)} *-1} + 3*\sqrt[3]{(7+\sqrt[2]{50)} *-1}

14 − 3 ∗ 7 + 50 2 3 − 3 ∗ 7 − 50 2 3 14 - 3*\sqrt[3]{7+\sqrt[2]{50} } - 3*\sqrt[3]{7-\sqrt[2]{50} }

14 − 3 ∗ ( 7 + 50 2 3 − 7 − 50 2 3 ) 14 - 3*(\sqrt[3]{7+\sqrt[2]{50} } - \sqrt[3]{7-\sqrt[2]{50} } )

From the main question :

X = 7 + 50 2 3 + 7 − 50 2 3 \large X = \sqrt[3]{7+\sqrt[2]{50}} + \sqrt[3]{7-\sqrt[2]{50}}

so

X 3 = 14 − 3 ∗ X X^3 = 14 - 3*X

we get :

X 3 + 3 X − 14 = 0 X^3 + 3X -14 = 0

so , the factors for 14 are : 1,2,7,14

then try 2

2 3 + 3 ∗ 2 − 14 = 0 2^3 + 3*2 -14 = 0 Already Done .

Finally , X = 2 X=2 Done .

3 Helpful 0 Interesting 0 Brilliant 0 Confused

@Mohammed Rizqallah , we really liked your comment, and have converted it into a solution. If you subscribe to this solution, you will receive notifications about future comments.

Brilliant Mathematics Staff - 3 years, 5 months ago

Log in to reply

@Brilliant Mathematics sorry, what did you mean about " converted it into a sol " so I write it as " sol " maybe I didn't get what you mean . Sorry I'm still new here and I have something prevents me to use a premium so I still learn use this website . Thanks for your advise .

Mohammed Rizqallah - 3 years, 5 months ago

Log in to reply

Hi Mohammed, you previously posted your solution as a report. We have copied what you've written in the report section as a solution.

Brilliant Mathematics Staff - 3 years, 5 months ago

Log in to reply

@Brilliant Mathematics – @Brilliant Mathematics , Aha I got it , Thanks alot .

Mohammed Rizqallah - 3 years, 5 months ago

0 pending reports

×

Problem Loading...

Note Loading...

Set Loading...